The work done by the gas when the volume increases from 12.0 m³ to 16.2 m³ at atmospheric pressure is 425.565 kJ. The change in internal energy of the gas when 254 kcal of heat is added is 636.931 kJ.
Explanation:To calculate the work done by the gas during a quasi-static expansion, we can use the formula W = P ΔV, where W is work, P is pressure, and ΔV is the change in volume. Given the atmospheric pressure is 1 atm or 101,325 Pa, and the volume change is from 12.0 m³ to 16.2 m³, the work done by the gas can be calculated as:
W = P ΔV = 101,325 Pa × (16.2 m³ - 12.0 m³)
W = 101,325 Pa × 4.2 m³
W = 425,565 J or 425.565 kJ
To calculate the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states that ΔU = Q - W, where Q is the heat added to the system. The heat is given as 254 kcal, which needs to be converted to joules (1 kcal = 4.184 kJ).
ΔU = Q - W = (254 kcal × 4.184 kJ/kcal) - 425.565 kJ
ΔU = 1062.496 kJ - 425.565 kJ
ΔU = 636.931 kJ
Learn more about Thermodynamics here:https://brainly.com/question/35546325
#SPJ12
A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop is:
Answer:
To oppose applied magnetic field current will flow in anticlockwise direction
Explanation:
Given:
Uniform magnetic field directed vertically upward.
Current will flow in clockwise direction
Here loop of wire is suddenly contracts to half so diameter of loop is reduced.
Hence less number of magnetic field line pass through the loop. This change in magnetic field lines lead to flow of current.
Now from lenz law flow of induced current will oppose the cause of its production
Therefore, to oppose applied magnetic field current will flow in anticlockwise direction.
Ms. Baudino wants to challenge her AP Physics students to use high order thinking and problem solving skills while engaged in a meaningful, real-world learning activity. What software can provide her students with this type of learning environmentA. Presentation
B. Word processing
C. Gaming and simulation
D. Presentation
Answer:
C. Gaming and simulation
Explanation:
In a presentation, you are just going to read from something and resume that to present. So there is not much higher order thinking or problem solving.
Word processing you will just be writing a test, so not much to learn in these aspects.
However
In gaming, there are different strategies your team may assume(which means that the students have to use high order thinking to choose the best strategy), you have to work well as a group, depending on the game(which is a needed skill on the real-world), and other things.
So the correct answer is:
C. Gaming and simulation
Answer:
C. Gaming and Simulation
Explanation:
Problem solving and high order thinking refers to the ability to use knowledge, facts, and data to effectively solve problems. This means the ability to assess problems and find suitable solutions. It is the ability to develop a well thought out solution within a reasonable time frame.
Integrating games and simulation into the learning process has a positive effect in developing the cognitive and behavioral abilities of students. Simulations promote the use of critical and evaluative thinking. Because they are ambiguous or open-ended, they encourage students to contemplate the implications of a scenario. It allows the teacher to properly engage the students while assuming a real world situation.
Due to constant practice, it allows the student to grasp the concept of a subject matter. The students are placed in a world defined by the teacher and they have enough motivation to learn because they now catch fun with activities that involve critical thinking.
Other options stated such as presentation and word processing only helps the students to give reports of activities and researches carried out. Gaming and simulation are the only ones that develop the critical thinking and problem solving skills of the students.
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor.
Answer:
each resistor is 540 Ω
Explanation:
Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance [tex]R_e[/tex] defined by the formula:
[tex]\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}[/tex]
Therefore, R/3 is the equivalent resistance of the initial circuit.
In the second circuit, two of the resistors are in parallel, so they are equivalent to:
[tex]\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\[/tex]
and when this is combined with the third resistor in series, the equivalent resistance ([tex]R''_e[/tex]) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):
[tex]R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}[/tex]
The problem states that the difference between the equivalent resistances in both circuits is given by:
[tex]R''_e=R_e+630 \,\Omega[/tex]
so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:
[tex]\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega[/tex]
A television meteorologist is able to inform viewers about the intensity of an approaching hurricane. However, the meteorologist can report the intensity of a tornado only after it has occurred. Why is this the case?
Answer: The meteorologist can report the intensity of a tornado only after it has occurred because (Tornado intensity is based on damage done.)
Explanation:
HURRICANE is a type of violent storm called topical cyclone which is characterised by a large rotating storm with high speed winds that forms over warm waters in tropical areas. They are form over the warm ocean water of the tropics. When warm moist air over the water rises, it is replaced by cooler air. The cooler air will then warm and start to rise. This cycle causes huge storm clouds to form. These storm clouds will begin to rotate with the spin of the Earth forming an organized system. If there is enough warm water, the cycle will continue and the storm clouds and wind speeds will grow causing a hurricane to form.
Hurricane can cause alot of damages such as flooding and storm surge. And they are also capable of developing to TORNADO.
The intensity of any tornado that occurred can only be measured through the damage it causes. This is because the correct intensity can only be determined while on site or remote sensing of the tornado which is quite impractical for wire scale use, therefore damages as a result of the tornado is used to quantify it.
A television meteorologist can inform about the approaching hurricane but not a tornado because of its formation and movements.
Meteorologist reportsA television meteorologist can inform about the approaching hurricane because it has t its pattern, which continues from the oceans and when it enters the land.
Meteorologist focuses their attention on these patterns for hurricanes in the Atlantic Ocean, and most of the time, their movement is predictable.
Tornados happen when the air pressure is strong in the ground for creating it.
The formation of a tornado can be sudden within minutes.
The direction of a tornado movement is unpredictable and based on its air pressure.
We have also seen meteorologists chasing a tornado and getting caught in it.
Therefore we can conclude that both have different patterns based on climate and temperature.
Learn more about Tornado here:
brainly.com/question/2826797
In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mass M. You shoot the striker with a speed V at the queen (which is at the center of the board) directly parallel to the rail that runs along the xaxis, hoping to pocket the queen in the bottom right corner. Assume the carrom board has edges that lay along the x- and y-axes and the board itself is in quadrant I and that the collision is elastic. If you make the shot, what is the relative velocity of the queen with respect to the striker after the collision?
Answer:
- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)
- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)
Hence, the relative velocity of the queen with respect to the striker after collision
= (3V/2) - (V/2)
= V m/s.
Explanation:
This is a conservation of Momentum problem.
Momentum before collision = Momentum after collision.
The mass of the striker = M
Initial Velocity of the striker = V (+x-axis)
Let the final velocity of the striker be u
Mass of the queen = (M/3)
Initial velocity of the queen = 0 (since the queen was initially at rest)
Final velocity of the queen be v
Collision is elastic, So, momentum and kinetic energy are conserved.
Momentum before collision = (M)(V) + 0 = (MV) kgm/s
Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)
Momentum before collision = Momentum after collision.
MV = Mu + (Mv/3)
V = u + (v/3)
u = V - (v/3) (eqn 1)
Kinetic energy balance
Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)
Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)
Kinetic energy before collision = Kinetic energy after collision
(MV²/2) = (Mu²/2) + (Mv²/6)
V² = u² + (v²/3) (eqn 2)
Recall eqn 1, u = V - (v/3); eqn 2 becomes
V² = [V - (v/3)]² + (v²/3)
V² = V² - (2Vv/3) + (v²/9) + (v²/3)
(4v²/9) = (2Vv/3)
v² = (2Vv/3) × (9/4)
v² = (3Vv/2)
v = (3V/2)
Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.
The final velocity of the striker after collision
= u = V - (v/3) = V - (V/2) = (V/2)
The relative velocity of the queen withrespect to the striker after collision
= (velocity of queen after collision) - (velocity of striker after collision)
= v - u
= (3V/2) - (V/2) = V m/s.
Hope this Helps!!!!
Answer:
The relative velocity of the queen is -vy
Explanation:
If the collision is elastic, thus e = 1. The expression is equal:
[tex]e=\frac{relative-velocity-of-approach}{relative-velocity-of-separation} \\relative-velocity-of-approach=relative-velocity-of-separation[/tex]
The relative velocity of separation is:
relative velocity of separation = 0 - vy = -vy
This expression means that:
velocity of queen - velocity of strikes = -vy
Thus the relative velocity of the queen with respect to the striker is equal to -vy
A bicycle tire is spinning clockwise at 2.50 rad/s. During a time period Dt 5 1.25 s, the tire is stopped and spun in the oppo- site (counterclockwise) direction, also at 2.50 rad/s. Calculate (a) the change in the tire’s angular velocity Dv and (b) the tire’s average angular acceleration aav .
Answer:
a) The change in the tire’s angular velocity is 5 rads/sec
b) The tire’s average angular acceleration is 4 rad/s²
Explanation:
initial angular speed , ω (initial) = - 2.5 rad/s
final speed , ω (final) = 2.5 rad/s
time = 1.25 s
a) The change in tire's angular speed
ω = ω (final) - ω (initial)
ω = 2.5 - ( -2.5) rad/s
ω = 2.5 + 2.5 (rad/s)
ω = 5 rad/s
b) The average angular acceleration
a = ω ÷ t
a = 5 rad/s ÷ 1.25 s
a = 4 rad/s²
a) The change in the tire’s angular velocity is 5 rads/sec
b) The tire’s average angular acceleration is 4 rad/s
initial angular speed , ω (initial) = - 2.5 rad/s
final speed , ω (final) = 2.5 rad/s
time = 1.25 s
a) The change in tire's angular speed
ω = ω (final) - ω (initial)
ω = 2.5 - ( -2.5) rad/s
ω = 2.5 + 2.5 (rad/s)
ω = 5 rad/s
b) The average angular acceleration
a = ω ÷ t
a = 5 rad/s ÷ 1.25 s
a = 4 rad/s²
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.25 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.540 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.51 m and t = 0.150 s .
Answer:
The transverse displacement is [tex]y(1.51 , 0.150) = 0.055 m[/tex]
Explanation:
From the question we are told that
The generally equation for the mechanical wave is
[tex]y(x,t) = Acos (kx -wt)[/tex]
The speed of the transverse wave is [tex]v = 8.25 \ m/s[/tex]
The amplitude of the transverse wave is [tex]A = 5.50 *10^{-2} m[/tex]
The wavelength of the transverse wave is [tex]\lambda = 0540 m[/tex]
At t= 0.150s , x = 1.51 m
The angular frequency of the wave is mathematically represented as
[tex]w = vk[/tex]
Substituting values
[tex]w = 8.25 * 11.64[/tex]
[tex]w = 96.03 \ rad/s[/tex]
The propagation constant k is mathematically represented as
[tex]k = \frac{2 \pi}{\lambda}[/tex]
Substituting values
[tex]k = \frac{2 * 3.142}{0. 540}[/tex]
[tex]k =11.64 m^{-1}[/tex]
Substituting values into the equation for mechanical waves
[tex]y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))[/tex]
[tex]y(1.51 , 0.150) = 0.055 m[/tex]
An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnetic field of 0.4 T at right angles to the direction of the electron’s motion. The mass of the electron is 9.11 × 10−31 kg and its charge is 1.60218 × 10−19 C. What is the magnitude of the force on the electron due to the magnetic field?
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
Light of wavelength 578.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 62.5 cm from the slit. The distance on the screen between the third order minimum and the central maximum is 1.35 cm . What is the width a of the slit in micrometers (μm)?
Answer:
[tex]80.6\mu m[/tex]
Explanation:
When light passes through a narrow slit, it produces a diffraction pattern on a distant screen, consisting of several bright fringes (constructive interference) alternated with dark fringes (destructive interference).
The formula to calculate the position of the m-th maximum in the diffraction pattern produced in the screen is:
[tex]y=\frac{m\lambda D}{d}[/tex]
where
y is the distance of the m-th maximum from the central maximum (m = 0)
[tex]\lambda[/tex] is the wavelength of light used
D is the distance of the screen from the slit
d is the width of the slit
In this problem, we have:
[tex]\lambda=578.0 nm = 578\cdot 10^{-9} m[/tex] is the wavelength
D = 62.5 cm = 0.625 m is the distance of the screen
We know that the distance between the third order maximum (m=3) and the central maximum is 1.35 cm (0.0135 m), which means that
[tex]y_3 = 0.0135 m[/tex]
For
m = 3
Therefore, rearranging the equation for d, we find the width of the slit:
[tex]d=\frac{m\lambda D}{y_3}=\frac{(3)(578\cdot 10^{-9})(0.625)}{0.0135}=80.3\cdot 10^{-6} m=80.6\mu m[/tex]
The width a of the slit will be "80.6 μm". To understand the calculation, check below.
Distance and WavelengthAccording to the question,
Light's wavelength, λ = 578.0 nm
Screen's distance, D = 62.5 or,
= 0.625 m
Third order maximum, m = 3
Central maximum = 1.35 cm or,
= 0.0135 m
We know the relation,
→ y = [tex]\frac{m \lambda D}{d}[/tex]
or,
→ Width, d = [tex]\frac{m \lambda D}{y_3}[/tex]
By substituting the values,
= [tex]\frac{3\times 578.10^{-9}\times 0.625}{0.0135}[/tex]
= 80.3 × 10⁻⁶ m or,
= 80.6 μm
Thus the above answer is correct.
Find out more information about wavelength here:
https://brainly.com/question/21630569
A large, open at the top, upright cylindrical tank contains fresh water with a density of 1.00 ✕ 103 kg/m3. If the air pressure is 101.3 kPa, determine the absolute pressure (in Pa) in the fluid at a depth of 29.0 m. Pa (b) Determine the force (in N) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 3.90 cm. N
To calculate the absolute pressure at a depth in a fluid, sum the atmospheric pressure and the pressure due to the fluid, while considering the fluid's density, gravitational acceleration, and depth. To find the force exerted by the fluid on a circular window at that depth, multiply the pressure by the area of the window.
Explanation:To begin with, you need to realize that the absolute pressure at a certain depth in a fluid is the sum of the atmospheric pressure and the pressure due to the fluid itself. This is because of Pascal's principle. In this case, the absolute pressure (Pabs) can be calculated as follows:
Pabs = Patm + density * g * h (height or depth in the fluid)
Here, Patm (atmospheric pressure) = 101.3 kPa = 101300 Pa (since 1 kPa = 1000 Pa), the density of water = 1.00 ✕ 103 kg/m3, acceleration due to gravity (g) = 9.81 m/s2 (approx), and h = 29.0 m
Using these values in the formula gives the absolute pressure in the fluid at a depth of 29.0 m:
Next, to determine the force exerted by the fluid on the circular window of the instrument probe at this depth, we have to understand that this force is actually the pressure at that depth multiplied by the area over which the pressure is exerted (Force = Pressure * Area) and the area of a circle = π * (d/2)²
Plug the determined pressure and given diameter into this equation to get the force in Newtons (N). Here, d = 3.90 cm = 0.039 m (since 1 cm = 0.01 m).
Learn more about Fluid Pressure Calculation here:https://brainly.com/question/32326195
#SPJ12
The absolute pressure in the fluid at a depth of 29.0 m is 2.83 × 10^4 Pa. The force exerted by the fluid on the window of the instrument probe at this depth is 10.9 N.
Explanation:To determine the absolute pressure in the fluid at a depth of 29.0 m, we need to consider the hydrostatic pressure. The hydrostatic pressure is given by the equation p = hρg, where p is the pressure, h is the depth, ρ is the density of the fluid, and g is the acceleration due to gravity. In this case, the density of the water is 1.00 × 103 kg/m3. We can substitute these values into the equation and solve for the pressure:
p = (29.0 m)(1.00 × 103 kg/m3)(9.81 m/s2) = 2.83 × 104 Pa
Therefore, the absolute pressure in the fluid at a depth of 29.0 m is 2.83 × 104 Pa.
To determine the force exerted by the fluid on the window of an instrument probe at this depth, we can use the formula F = PA, where F is the force, P is the pressure, and A is the area. The area of the circular window can be calculated using the formula A = πr2, where r is the radius of the window. Given that the diameter of the window is 3.90 cm, the radius is half of the diameter, so r = 1.95 cm = 0.0195 m. Substituting the values into the formulas, we can find the force:
F = (2.83 × 104 Pa)(π(0.0195 m)2) = 10.9 N
Therefore, the force exerted by the fluid on the window of the instrument probe at a depth of 29.0 m is 10.9 N.
Learn more about Pressure in Fluids here:https://brainly.com/question/29039234
#SPJ11
For the questions below, answer if the statement is true or false and explain how you reach to your conclusion. a. The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body. b. The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs. c. Mass, energy, and temperature are the examples of intensive properties. d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is COPHP = COPR + 1. e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11. f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%. g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.
Answer:
a) Seems true at first glance, but on further inspection, the statement is false.
b) True
c) False
d) True
e) True
f) False
g) False
Explanation:
Taking the statements one by one
a.) The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body.
The Clausius Statement denies the possibility of heat transfer from a cooler to a hotter body without extra work. It does not outrightly state that there is no possibility of heat transfer from a cooler to a hotter body.
For example, an Air conditioner or refrigerator rejects heat from a cold reservoir to a hot reservoir.
So, this statement is false.
b) The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs.
The Coefficient of Performance of a reversible cycle is the maximum efficiency possible. It is the efficiency of a Carnot Engine.
Hence, it is greater than or equal to the Coefficient of Performance of an irreversible cycle.
COP(reversible) ≥ COP(irreversible)
This statement is true.
c. Mass, energy, and temperature are the examples of intensive properties.
Intensive properties are properties of thermodynamic systems that do not depend on the extent of the system. They are the same for a particular size of substance and stay the same if the size of the substance is doubled or halved. Examples include temperature, specific capacity, specific volume, every specific property basically, etc.
Extensive properties depend on the extent of the system. They double or half when the size of the extent doubles or halves respectively.
Mass and Energy are Extensive properties.
Temperature is the only intensive property among these options.
This statement is false.
d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is
COPHP = COPR + 1.
The coefficients of performance for reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs are indeed related through
COP(HP) = COP(R) + 1
COP of a heat pump = COP(HP) = (Qh/W)
COP of a refrigerator = COP(R) = (Qc/W)
But, Qh = Qc + W
Divide through by W
(Qh/W) = (Qc/W) + (W/W)
COP(HP) = COP(R) + 1 (Proved!)
This Statement is true.
e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11.
The COP is given as (1/efficiency).
Efficiency = 1 - (Tc/Th)
Tc = temperature of cold thermal reservoir in Kelvin = 350°C = 623.15 K
Th = temperature of hot thermal reservoir in Kelvin = 550°C = 823.15 K
Efficiency = 1 - (623.15/823.15)
= 1 - 0.757 = 0.243
COP = (1/Efficiency) = (1/0.243) = 4.11
This statement is true.
f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%.
In the absence of friction and other irreversibilities and for a heat engine to have 100% efficiency, the temperature of its cold reservoir has to be 0 K or the tempersture of its hot reservoir has to be infinity.
Efficiency = 1 - (Tc/Th)
For efficiency to be 1,
(Tc/Th) = 0; that is, Tc = 0 or Th = infinity
These two aren't physically possible and for 100% efficiency to happen, the heat engine will have to violate the Kelvin-Planck's statement of the second law of thermodynamics.
According to Kelvin-Planck's statement of the second law of thermodynamics, net amount of work cannot be produced by exchanging heat with single reservoir i.e. there will be another reservoir to reject heat.
Hence, a heat engine cannot have an efficiency of 100%.
This statement is false.
g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.
The COP of a refrigerator is given as
COP = (Qcold)/W
Qcold = Heat rejected from the cold reservoir = Heat rejected from refrigerated space = 60 KJ/min
W = work done on the system = electrical power consumed = 50 KJ/min
COP = (60/50) = 1.2
This system does not violate the first law of thermodynamics.
This statement is false.
Hope this Helps!!!
A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides 2.0 m before stopping. Use work and energy to find the suitcase’s coefficient of kinetic friction on the floor.
Answer:
0.0367
Explanation:
The loss in kinetic energy results into work done by friction.
Since kinetic energy is given by
KE=0.5mv^{2}
Work done by friction is given as
W= umgd
Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.
Making u the subject of the formula then we deduce that
[tex]u=\frac {v^{2}}{2gd}[/tex]
Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then
[tex]u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367[/tex]
Therefore, the coefficient of kinetic friction is approximately 0.0367
The coefficient of kinetic friction on the floor is approximately 0.037. This is found by equating the work done by friction, which is equal to the change in kinetic energy of the suitcase, to the frictional force times the distance over which it acts.
Explanation:To solve this problem we need to understand that the work done by friction is equal to the change in kinetic energy of the suitcase, as per the work-energy theory. The initial kinetic energy of the suitcase is given by (1/2)*m*v^2, where m is the mass (15 kg) and v is the velocity (1.2 m/s). The final kinetic energy is 0 because the suitcase comes to a stop. Thus, the work done by friction is equal to the initial kinetic energy.
The work done by friction can also be expressed as the force of friction times the distance over which it acts, which in this case, is the distance the suitcase traveled (2.0 m). We know that the frictional force is equal to the product of the coefficient of kinetic friction and the normal force. Here, the normal force is equal to the weight of the suitcase, which is weight = m*g, where g is the gravitational acceleration (9.8 m/s^2).
Setting the two expressions for the work done by friction equal to each other gives us the equation (1/2)*m*v^2 = u*m*g*d. Solving this equation for u (the coefficient of kinetic friction) gives us: u = [(1/2)*v^2] / (g*d).
Substituting the given values into this equation, we find that the coefficient of kinetic friction on the floor is approximately 0.037.
Learn more about Coefficient of kinetic friction here:https://brainly.com/question/30394536
#SPJ3
Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable shearing force in each nail is 740 N, determine the largest permissible spacing s between the nails.
Answer:
100.49 mm
Explanation:
Given that :
The vertical shear force = 1200 N
Allowable shearing force in each nail = 740 N
From the diagram attached below, We first determine the horizontal force per unit length on the lower surface of the upper flange
[tex]q = \frac{VQ_{1-1}}{I_{NA}} \ \ \ N/mm[/tex]
where;
[tex]I_{NA} = \frac{50*100^3}{12}+2[\frac{150*50^3}{12}+ (150*50)*( \bar {y}^2][/tex]
[tex]I_{NA} = \frac{50*100^3}{12}+2[\frac{150*50^3}{12}+ (150*50)*( 75)^2][/tex]
[tex]I_{NA} = 91.667*10^6 \ mm^4[/tex]
Also;
[tex]Q_{1-1} = A \bar{y}[/tex]
where A = area above (1-1)
[tex]Q_{1-1} = 50*150*75[/tex]
[tex]Q_{1-1} =56.25*10^4 \ \ \ mm^3[/tex]
Therefore ;
[tex]q = \frac{1220*56.25*10^4}{91.667*10^6}[/tex]
[tex]q = 7.3636 \ N/mm[/tex]
Now; the largest permissible spacing s between the nails. [tex]S_{max} = \frac{740}{7.3636} = 100.49 \ \ \ mm[/tex]
To determine the largest permissible spacing between the nails in a beam, calculate the total shearing force the nails can withstand and divide it by the area of contact. The largest permissible spacing is 14.8 cm.
Explanation:To determine the largest permissible spacing between the nails, we need to calculate the total shearing force that the beam can withstand. Since each nail can withstand a maximum shearing force of 740 N, the total shearing force that the nails can withstand is 3 multiplied by 740 N, which equals 2220 N. We can set up an equation to find the largest permissible spacing, s, by dividing the total shearing force by the area of contact between the nails and the beam. The area of contact can be calculated by multiplying the thickness of the beam by the length of the nails, so it is 3 multiplied by 50 mm (or 0.05 m) multiplied by s. So, we have the equation:
2220 N = 0.05 m x 3 x s
Simplifying the equation, we get
2220 N = 0.15 m x s
Dividing both sides of the equation by 0.15 m, we find:
s = 2220 N / 0.15 m
s = 14800 mm (or 14.8 cm)
Therefore, the largest permissible spacing between the nails is 14.8 cm.
Learn more about Shearing force here:https://brainly.com/question/30763282
#SPJ11
A tire has a tread pattern with a crevice every 2.00 cm. The repetitive striking of the crevice edges on the road results in a vibration as the tire moves. What is the frequency of these vibrations if the car moves at 14.4 m/s?
Answer:
Frequency of vibration will be equal to 720 Hz
Explanation:
It is given that car moves with a speed of 14.4 m/sec
Its mean that car cover 14.4 m in 1 sec
We have to find the frequency of vibration corresponding to this velocity
It is given that car has a tread pattern with a crevice every 2 cm
So number of crevices moved in 1 sec will be equal to [tex]=\frac{14.4}{0.02}=720[/tex]
So frequency of these vibration will be equal to 720 Hz
A sphere moves in simple harmonic motion with a frequency of 4.80 Hz and an amplitude of 3.40 cm. (a) Through what total distance (in cm) does the sphere move during one cycle of its motion? cm (b) What is its maximum speed (in cm/s)? cm/s Where does this maximum speed occur? as the sphere passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these (c) What is the maximum magnitude of acceleration (in m/s2) of the sphere? m/s2 Where in the motion does the maximum acceleration occur? as the sphere passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these
Answer:
a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b) The maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
The maximum acceleration occurs at maximum excursion from equilibrium.
Explanation:
Given that :
Frequency (f) = 4.80 Hz
Amplitude (A) = 3.40 cm
a)
The total distance traveled by the sphere during one cycle of simple harmonic motion is:
d = 4A (where A is the Amplitude)
d = 4(3.40 cm)
d = 13.60 cm
Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b)
As we all know that:
[tex]x = Asin \omega t[/tex]
Differentiating the above expression with respect to x ; we have :
[tex]\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)[/tex]
[tex]v = A \omega cos \omega t[/tex]
Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;
Then:
[tex]v_{max} = A \omega[/tex]
We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium
substituting [tex]2 \pi f[/tex] for [tex]\omega[/tex] in the above expression;
[tex]v_{max} = A(2 \pi f)[/tex]
[tex]v_{max} = 3.40 cm (2 \pi *4.80)[/tex]
[tex]v_{max} = 102.54 \ cm/s[/tex]
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
[tex]v = A \omega cos \omega t[/tex]
By differentiation with respect to t;
[tex]\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)[/tex]
[tex]a =- A \omega^2 sin \omega t[/tex]
The maximum acceleration of the sphere is;
[tex]a_{max} =A \omega^2[/tex]
where;
[tex]w = 2 \pi f[/tex]
[tex]a_{max} = A(2 \pi f)^2[/tex]
where A= 3.40 cm = 0.034 m
[tex]a_{max} = 0.034*(2 \pi *4.80)^2[/tex]
[tex]a_{max} = 30.93 \ m/s^2[/tex]
The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of [tex]x = \pm A[/tex]
The sphere moves a total distance of 6.80 cm in one cycle. Its maximum speed is 102.5 cm/s as it passes through equilibrium and the maximum acceleration is at maximum excursion from equilibrium, with a magnitude of 18.5 m/s².
Explanation:The sphere's total distance moved in one cycle is twice the amplitude, as it moves to the maximum amplitude and back again. Therefore, in the case of this sphere, it moves through a total distance of 6.80 cm during one cycle. The maximum speed of an object in simple harmonic motion occurs as it passes through equilibrium. You can calculate this speed using the formula v = ωA, where ω is the angular frequency and A is the amplitude.
The angular frequency is ω = 2πf, where f is the frequency. Thus, the maximum speed of the sphere is approximately 102.5 cm/s. Lastly, the maximum magnitude of acceleration occurs at maximum excursion from equilibrium, and can be calculated with the formula derived from Newton's second law a = ω²A, thus the acceleration is 18.5 m/s².
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
#SPJ3
A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.240 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.70 s. What average emf is induced in the second coil if it has a diameter of 3.90 cm and N = 48? Express your answer in microvolts to two significant figures.What is the induced emf if the diameter is 7.80 cm and N = 96? Express your answer in volts to two significant figures.
Answer:
Induced emf in first coil is 0.986 [tex]\mu T[/tex] and in second case 0.396 [tex]\mu T[/tex]
Explanation:
Given:
Number of turns per centimeter [tex]n = 40[/tex]
Current [tex]I = 0.240[/tex] A
Current rate [tex]\frac{dI}{dt} = \frac{0.240}{0.70} = 0.343[/tex] [tex]\frac{A}{s}[/tex]
The magnetic field in solenoid is given by,
[tex]B = \mu _{o} nI[/tex]
Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex]
We write,
[tex]\frac{dB}{dt} = \mu_{o} n \frac{dI}{dt}[/tex]
[tex]\frac{dB}{dt} = 4\pi \times 10^{-7} \times 40 \times 0.343[/tex]
[tex]\frac{dB}{dt} = 172.3 \times 10^{-7}[/tex]
(A)
Number of turns [tex]N = 48[/tex]
Radius of coil [tex]r = \frac{d}{2} = 1.95 \times 10^{-2}[/tex] m
From faraday's law
[tex]\epsilon = NA \frac{dB}{dt}[/tex]
Where [tex]A = \pi r^{2} = 3.14 (1.95 \times 10^{-2} ) ^{2} = 11.93 \times 10^{-4}[/tex] [tex]m^{2}[/tex]
[tex]\epsilon = 48 \times 11.93 \times 10^{-4} \times 172.3 \times 10^{-7}[/tex]
[tex]\epsilon = 98665.87 \times 10^{-11}[/tex]
[tex]\epsilon = 0.986 \mu T[/tex]
(B)
Number of turns [tex]N = 96[/tex]
Radius of coil [tex]r = \frac{d}{2} = 3.9 \times 10^{-2}[/tex] m
From faraday's law
[tex]\epsilon = NA \frac{dB}{dt}[/tex]
Where [tex]A = \pi r^{2} = 3.14 (3.9 \times 10^{-2} ) ^{2} = 47.76 \times 10^{-4}[/tex] [tex]m^{2}[/tex]
[tex]\epsilon = 48 \times 47.96 \times 10^{-4} \times 172.3 \times 10^{-7}[/tex]
[tex]\epsilon = 396648.38 \times 10^{-11}[/tex]
[tex]\epsilon = 0.396 \mu T[/tex]
Therefore, induced emf in first coil is 0.986 [tex]\mu T[/tex] and in second case 0.396 [tex]\mu T[/tex]
Light of wavelength 614 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?
Answer:
The least and second least thicknesses of the film are 0.115 um and 0.346 um respectively.
Explanation:
Optical path length ===> 2n * t = (m + 0.5) * λ
λ = 614 nm , n = 1.33
Substitute in the parameters in the equation.
∴ 2(1.33) * t = (m + 0.5) * 614
2.66 * t = 614m + 307
t = (614m + 307) / 2.66 .............(1)
(a) for m = 0
t = (614m + 307) / 2.66
t = (614(0) + 307) / 2.66
t = 307 / 2.66
t = 115 nm == 0.115 um
(b) for m = 1
t = (614(1) + 307) / 2.66
t = (614 + 307) / 2.66
t = 921 / 2.66
t = 346.24 nm = 0.346 um
Energy transformations worksheet I need help!
The answers of the all the questions are given as 1-a,2-b,3-b,4-c,5-c,6-a,7-b,8-c,9-b,10-b or c,11-b,
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
A whole branch of physics, thermodynamics, deals with how heat is transferred between different systems and how work is done in the process.
Almost every transfer of energy that takes place in real-world physical systems does so with an efficiency of less than 100% and results in some thermal energy.
This energy is usually in the form of low-level thermal energy. Here, low-level means that the temperature associated with the thermal energy is close to that of the environment.
It is only possible to extract work when there is a temperature difference, so low-level thermal energy represents 'the end of the road' of energy transfer. No further useful work is possible; the energy is now 'lost to the environment.
To know more about thermal energy follow
https://brainly.com/question/19666326
There is a current I flowing in a clockwise direction in a square loop of wire that is in the plane of the paper. If the magnetic field B is toward the right, and if each side of the loop has length L, then the net magnetic force acting on the loop is:
a. 2ILB
b. ILB
c. IBL2
d. Zero
Answer:
(d) Zero
Explanation:
Given:
Side of the square, = L
Magnetic field, = B
According to Faraday, the net force acting on a conductor is equal to product of magnetic field, current, length of conductor and sine of the angle between the field and current.
F = BILsinΦ
Where:
B is magnetic field
I is current in the loop
L is length of the loop
Φ is the angle between the magnetic field and current
Φ = 0, since the current and magnetic field are directed in opposite directions.
F = BILsin(0)
F = 0
Therefore, the net magnetic force acting on the loop is zero.
The correct answer is (d) Zero
The correct option is Option d (zero). The net magnetic force acting on the loop is zero because the forces on the sides of the loop cancel each other out. Thus, the correct answer is d. Zero.
In a uniform magnetic field, the net force on a current-carrying loop in a magnetic field can be determined using the equation F = I \times B. We need to apply this equation to each side of the loop.
Step-by-Step Explanation:
For the two sides parallel to the magnetic field, the forces are zero because the angle between the current direction and the magnetic field is zero.For the two vertical sides perpendicular to the magnetic field: one side carries current downward, and the other side carries current upward. Applying the right-hand rule shows that the forces on these sides are equal in magnitude but opposite in direction.Since these opposite forces are of equal magnitude, they cancel each other out.Therefore, the net magnetic force acting on the loop is zero, making the correct answer d. Zero.
The hill is covered in gravel so that the truck's wheels will slide up the hill instead of rolling up the hill. The coefficient of kinetic friction between the tires and the gravel is k. This design has a spring at the top of the ramp that will help to stop the trucks. This spring is located at height h. The spring will compress until the truck stops, and then a latch will keep the spring from decompressing (stretching back out). The spring can compress a maximum distance x because of the latching mechanism. Your job is to determine how strong the spring must be. In other words, you need to find the spring constant so that a truck of mass mt, moving at an initial speed of v0, will be stopped. For this problem, it is easiest to define the system such that it contains everything: Earth, hill, truck, gravel, spring, etc. In all of the following questions, the initial configuration is the truck moving with a speed of v0 on the level ground, and the final configuration is the truck stopped on the hill with the spring compressed by an amount x. The truck is still in contact with the spring. Solve all of the questions algebraically first. Then use the following values to get a number for the desired answer.
Answer:
Explanation:
check attachment for the solution.
(a) The network done is equal to zero.
(b) The change in the potential energy is equal to the 5.54 ×10⁶ J
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
The network done is given as;
[tex]\rm W= Fd cos \alpha \\\\ \rm W= F \times D cos 90^0 \\\\\rm W=0\ J[/tex]
Hence the net work done is equal to zero.
(b) The change in the potential energy is equal to the 5.54 ×10⁶ J
[tex]\rm \triangle U_g= mg (h+x sin\theta)\\\\ \rm \triangle U=1200 \times 9.81(45+3.5sin 37.8^0)\\\\ \rm \triangle U_g=5.5443 \times 10^6 \ J[/tex]
Hence the changes in the potential energy are equal to the 5.54 ×10⁶ J.
(c) The change in the thermal energy will be 3.3425×10⁶ J.
The formula for the thermal energy change is found as;
[tex]\rm \triangle E_{thermal} =\mu_k mgcos \theta (\frac{h-L}{sin \theta} +x )\\\\ \rm \triangle E_{thermal} =0.6 \times 1200 \times 9.81 \times 37.8^0 ( (\frac{45-18.4}{sin 37.8^0} +3.5 )\\\\ \rm \triangle E_{thermal} =3.342 \times 10^6 \ J[/tex]
Hence the changes in the thermal energy will be 3.3425×10⁶ J.
To learn more about the work done refer to the link ;
https://brainly.com/question/3902440
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Answer:
Coefficient of static friction will be equal to 0.642
Explanation:
We have given acceleration [tex]a=6.3m/sec^2[/tex]
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We have to find the coefficient of static friction between truck and a cabinet will
We know that acceleration is equal to [tex]a=\mu g[/tex], here [tex]\mu[/tex] is coefficient of static friction and g is acceleration due to gravity
So [tex]\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642[/tex]
So coefficient of static friction will be equal to 0.642
A 20 kg box on a horizontal frictionless surface is moving to the right at a speed of 4.0 m/s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.50 m, and the box then oscillates back and forth. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero. ____ Positive ____ Negative ____ Zero Justify your answer. ii. Calculate the magnitude of the work described in part i. (b) Calculate the spring constant of the spring. (c) Calculate the magnitude of the maximum acceleration of the box. (d) Calculate the frequency of the oscillation of the box. (e) Let x
a)
i) Negative
ii) 160 J
b) 1280 N/m
c) [tex]32 m/s^2[/tex]
d) 1.27 Hz
e)
i) See attached plot
ii) See attached plot
Explanation:
a)
i) The work done by a force is given by
[tex]W=Fx cos \theta[/tex]
where
F is the force
x is the displacement of the object
[tex]\theta[/tex] is the angle between the direction of the force and the direction
Here we have:
- The force that the spring exerts on the box is to the left (because the box is moving to the right, trying to compress the spring)
- The displacement of the box is to the right
So, F and x have opposite direction, and so [tex]\theta=180^{\circ}[/tex] and [tex]cos \theta=-1[/tex], which means that the work done is negative.
ii)
According to the work-energy theorem, the work done by the spring is equal to the change in kinetic energy of the box:
[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
[tex]K_i[/tex] is the initial kinetic energy of the box
[tex]K_f[/tex] is the final kinetic energy
m = 20 kg is the mass of the box
u = 4.0 m/s is its initial speed
v = 0 m/s is the final speed (the box comes to rest)
Therefore,
[tex]W=\frac{1}{2}(20)(0)^2-\frac{1}{2}(20)(4.0)^2=-160 J[/tex]
So, the magnitude is 160 J.
b)
The elastic energy stored in a spring when it is compressed is given by
[tex]U=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the stretching/compression of the spring
Due to the law of conservation of energy, the kinetic energy lost by the box is equal to the elastic energy gained by the spring, so:
[tex]|W|=U=\frac{1}{2}kx^2[/tex]
We have
[tex]|W|=160 J[/tex]
x = 0.50 m is the maximum compression of the spring
Solving for k:
[tex]k=\frac{2U}{x^2}=\frac{2(160)}{(0.50)^2}=1280 N/m[/tex]
c)
The magnitude of the force exerted on the box is given by
[tex]F=kx[/tex]
where
k = 1280 N/m is the spring constant
x = 0.50 m is the compression of the spring
Substituting,
[tex]F=(1280)(0.50)=640 N[/tex]
Now we can find the maxmum acceleration using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
F = 640 N is the maximum force
m = 20 kg is the mass of the box
So,
[tex]a=\frac{640}{20}=32 m/s^2[/tex]
d)
The frequency of oscillation of a spring-mass system is given by
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Here we have:
k = 1280 N/m is the spring constant of this spring
m = 20 kg is the mass of the box
So, the frequency of this system is:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{1280}{20}}=1.27 Hz[/tex]
e)
i)
Here we want to sketch the kinetic energy of the box as a function of the position, x: find this graph in attachment.
In a spring-mass oscillating system, the kinetic energy is zero when the system is at the extreme position, i.e. when the spring is maximum compressed/stretched. In this problem, this happens when x = - 0.50 m and x = +0.50 m (we called x = 0 the position of equilibrium of the spring). In these positions in fact, the mass has zero speed, so its kinetic energy is zero.
On the other hand, the box has maximum speed when x = 0 (because it's the moment where all the elastic energy is converted into kinetic energy, which is therefore maximum, and so the speed is also maximum).
ii)
Here we want to plot the acceleration of the box as a function of the position x: find the graph in attachment.
In a spring-mass system, the acceleration is proportional to the negative of the displacement, since the restoring force
[tex]F=-kx[/tex]
By rewriting the force using Newton's second Law, we have
[tex]ma=-kx \\a=-\frac{k}{m}x[/tex]
Which means that acceleration is proportional to the displacement, but with opposite sign: so, this graph is a straight line with negative slope.
A mass m at the end of a spring of spring constant k is undergoing simple harmonic oscillations with amplitude A. show answer No Attempt 33% Part (a) At what positive value of displacement x in terms of A is the potential energy 1/9 of the total mechanical energy? x = | α β π θ A B d g h j k m P S t ( ) 7 8 9 HOME ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. No Attempt No Attempt 33% Part (b) What fraction of the total mechanical energy is kinetic if the displacement is 1/2 the amplitude? No Attempt No Attempt 33% Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3? All content © 2020 Expert TA, LLC
Answer:
a) x = 0.33 A , b) K = ¾ Em , c) the kinetic energy increases 9 times
Explanation:
a) In a simple harmonic motion the mechanical energy is conserved and is expressed by the relation
Em = ½ k A²
At all points of movement the mechanical energy and
Em = K + U
remember that potential energy is
U = ½ k x²
they ask us for the point where
U = 1/9 Em
we substitute
½ k x² = 1/9 (½ k A²)
x = √1/9 A
x = 0.33 A
b) At all points the equation for mechanical energy is
Em = K + U
K = Em - U
K = ½ k A² - ½ k x²
K = ½ k (A² -x²)
at point x = ½ A
K = ½ k (A² - ¼ A²)
K = (½ k A²) ¾
K = ¾ Em
U = ½ k (½A) 2
U = ¼ Em
the fraction of energy e
U / K = 1/3
c) Kinetic energy is
K = ½ k v²
the system is described by the expression
x = A cos (wt + Ф)
speed is defined
v = dx / dt
v = - A w sin (wt + Ф)
we substitute
K = ½ k (- A w sin ( wt + Ф))²
We write this equation for the initial amplitude A
K₀ = ½ k (w sin (wt + Ф))² A²
we write it for the new amplitude A´ = 3 A
K = ½ k (w sin (wt +Ф))² (3A)²
the relationship between this energies is
K / K₀ = 9
whereby the kinetic energy increases 9 times
The displacement for 1/9 of mechanical energy is A/3. When the displacement is half amplitude, 3/4 of total mechanical energy becomes kinetic energy. If the amplitude increases by a factor of 3, the maximum kinetic energy increases by a factor of 9.
Explanation:In this context, we know that the total mechanical energy (E) of the system is constant and is the sum of kinetic and potential energy. The potential energy (U) of a mass-spring system is given by the formula U = 1/2 kx².
Part (a)
We want to find the displacement x such that the potential energy is 1/9 total mechanical energy, thus we set U = E/9. From substituting U = 1/2kx² and E = 1/2kA² in the equation,
we find that x²/A² = 1/9, which gives x = A/3.
Part (b)
We know that when the particle is at x = A/2 the potential energy is U = 1/2 k(A/2)² = 1/8 kA².
The total energy E = 1/2 kA², so the kinetic energy (K) is given by E - U = 1/2kA² - 1/8kA² = 3/8 kA².
The fraction of the total energy which is kinetic is (K/E) = (3/8)/(1/2) = 3/4.
Part (c)
The maximum kinetic energy is the total mechanical energy (since potential energy is zero at maximum kinetic energy), which is E = 1/2*kA².
If we triple the amplitude, the new energy E' = 1/2*k(3A)² = 9/2*kA².
The factor by which the maximum kinetic energy changes is (E'/E) = (9/2)/(1/2) = 9.
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
#SPJ3
A small sphere of mass m is launched horizontally over a body of water from a height h above the water and with a launch speed v0. Determine expressions for the following in terms of m, v0, h, and g. Air resistance is negligibly small.
(a) W is the amount work done by the force of gravity on the projectile during its flight.
(b) ΔKE is the change in kinetic energy ΔKE of the projectile from the time it was fired until it hits the water.
(c) KEf is the final kinetic energy KEf of the projectile as it hits the water.
The work done by the force of gravity on the projectile is given by W = mgh. The change in kinetic energy of the projectile is given by ΔKE = (1/2)mv0^2. The final kinetic energy of the projectile as it hits the water is given by KEf = (1/2)mv0^2.
Explanation:(a) Work done by the force of gravity:
The work done by the force of gravity on the projectile during its flight can be calculated using the formula:
W = mgh
where W is the work done, m is the mass of the sphere, g is the acceleration due to gravity, and h is the height above the water.
(b) Change in kinetic energy:
The change in kinetic energy of the projectile from the time it was fired until it hits the water can be calculated using the formula:
ΔKE = (1/2)mv0^2
where ΔKE is the change in kinetic energy, m is the mass of the sphere, and v0 is the launch speed.
(c) Final kinetic energy:
The final kinetic energy of the projectile as it hits the water can be calculated using the same formula as in part (b):
KEf = (1/2)mv0^2
Learn more about Work, Kinetic Energy here:https://brainly.com/question/32752469
#SPJ12
How much current does a 100-W lamp draw when connected to 120 V?
P/V = I
Answer:
0.83 amps is what I believe is the answer.
Light from a laser (lambda= 406.192 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 5.937 m behind the slits. 24 bright fringes are seen, spanning a distance of 39.835 mm (they are not necessarily centered, you don't know where the center is, this is not important here). What is the spacing (in mm) between the slits? Give the numerical answer only, in m
Answer:
The spacing between the slits is [tex]d = 0.00145m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 406.192nm = 406.192*10^{-9} m[/tex]
The distance of the slit from the screen is [tex]D = 5.937 \ m[/tex]
The number of bright fringe is [tex]n = 24[/tex]
The length the fringes span is [tex]L = 39.835 mm = \frac{39.835 }{1000} = 0.0398 m[/tex]
The fringe width (i.e the distance of between two successive bright or dark fringe) is mathematically represented as
[tex]\beta = \frac{\lambda D}{d}[/tex]
Where d is the distance between the slits
[tex]\beta[/tex] is the fringe width which can also be evaluated as
[tex]\beta = \frac{L}{n}[/tex]
Substituting values
[tex]\beta = \frac{0.0398}{24}[/tex]
[tex]\beta = 1.660 *10^{-3}[/tex]
Making d the subject of formula in the above equation
[tex]d = \frac{\lambda D}{\beta }[/tex]
Substituting values
[tex]d = \frac{406.192 *10^{-9} * 5.937 }{1.660 *10^{-3}}[/tex]
[tex]d = 0.00145m[/tex]
A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the plane by an ideal spring that is aligned with the surface and attached to a wall above the brick. The spring has a spring constant (force constant) of 120 N/m. By how much does the spring stretch with the brick attached?
Answer: 0.137 m
Explanation:
Given
Mass of brick, m = 3 kg
Angle of inclination, Φ = 34°
Force constant of the spring, k = 120 N/m
The force of the brick, F can be gotten using the relation
F = mg
F = 3 * 9.8
F = 29.4 N
Now, the force parallel to the incline, F(p) can be gotten using the formula,
F(p) = F sinΦ, so that
F(p) = 29.4 * sin 34
F(p) = 29.4 * 0.559
F(p) = 16.4 N
The stretch distance then is,
d = F(p) / k * 1 m
d = 16.4 / 120
d = 0.137 m
Thus, the spring stretched by a distance of 0.137 m
Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path in terms of the book’s mass mb , tangential speed vb , radius R of the path, and physical constants, as appropriate. Do not substitute any numerical values for variables or physical constants.
a)
i) 120 s
ii) 1.57 m/s
b)
i) See attachment
ii) Up
c) [tex]N=mg+m\frac{v_b^2}{R}[/tex]
d) Greater than
Explanation:
The problem is incomplete: find the complete text in attachments.
a)
i) The period of revolution of the book is equal to the total time taken by the book to complete one revolution.
Looking at the graph, the period of revolution can be estimated by evaluating the difference in time between two consecutive points of the motion of the book that have the same shape.
For instance, we can evaluate the period by calculating the difference in time between two consecutive crests. We see that:
- The first crest occur at t = 90 s
- The second crest occurs at t = 210 s
Therefore, the period of revolution is
T = 210 - 90 = 120 s
ii)
The tangential speed of the book is given by the ratio between the distance covered during one revolution (so, the perimeter of the wheel) and the period of revolution.
Mathematically:
[tex]v_b=\frac{2\pi R}{T}[/tex]
where
R is the radius of the wheel
T = 120 s is the period
From the graph, we see that the maximum position of the book is x = +30 m, while the minimum position is x = -30 m, so the diameter of the wheel is
d = +30 - (-30) = 60 m
So the radius is
R = d/2 = 30 m
So, the speed is
[tex]v_b=\frac{2\pi (30)}{120}=1.57 m/s[/tex]
b)
i) See in attachment the free-body diagram of the book at its lowest position.
There are 2 forces acting on the book at the lowest position:
- The weight of the book, of magnitude
[tex]W=mg[/tex]
where m is the mass of the book and g the acceleration due to gravity. This force acts downward
- The normal force exerted by the bench on the book, of magnitude N. This force acts upward
ii)
When the book is at its lowest position, it is moving horizontally at constant speed.
However, the book is accelerating. In fact, acceleration is the rate of change of velocity, and velocity is a vector, so it has both a speed and a direction; here the speed is not changing, however, the direction is changing (upward), so the book has an upward net acceleration.
According to Newton's second law of motion, the net vertical force on the book is proportional to its net vertical acceleration:
[tex]F=ma[/tex]
where F is the net force, m is the mass, a is the acceleration. Therefore, since a is different from zero, the book has a net vertical force, in the same direction of the acceleration (so, upward).
c)
As we said in part b), there are two forces acting on the book at its lowest position:
- The weight, [tex]W=mg[/tex], downward
- The normal force of the bench, N, upward
Since the book is in uniform circular motion, the net force on it must be equal to the centripetal force [tex]m\frac{v_b^2}{R}[/tex], so we can write:
[tex]N-mg=m\frac{v_b^2}{R}[/tex]
where
[tex]v_b[/tex] is the speed of the book
R is the radius of the path
Therefore, we find an expression for the normal force:
[tex]N=mg+m\frac{v_b^2}{R}[/tex]
d)
As we said in part c) and d):
- The normal force acting on the book at its lowest position is
[tex]N=mg+m\frac{v_b^2}{R}[/tex]
- The weight (force of gravity) of the book is
[tex]W=mg[/tex]
By comparing the two equations above, we observe that
[tex]N>W[/tex]
Therefore, we can conclude that the normal force exerted by the bench on the book is greater than the weight of the book.
A negative charge of - 4.0 x 10-5 C and a positive charge of
7.0 x 10 -5 C are separated by 0.15 m. What is the force
between the two charges?
Answer:
1117.51N/C
Explanation:
The magnitude of the electric force is given by:
[tex]|\vec{F}|=|k\frac{q_1q_2}{r^2}|[/tex]
k: Coulomb's constant = 8.98*10^9Nm^/2C^2
r: distance between the charges = 0.15m
By replacing the values of q1, q2, k and r you obtain:
[tex]|\vec{F}|=|(8.98*10^9Nm^2/C^2)\frac{(-4.0*10^{-5}C)(7.0*10^{-5}C)}{(0.15m)^2}|=1117.51\frac{N}{C}[/tex]
hence, the force between the charges is 1117.51N/C
Answer:
The force of attraction between the charges is 1120 N
Explanation:
Given:
positive charge of the particle, q1 = 7 x 10^-5 C
negative charge of the particle, q2 = -4 x 10^-5 C
distance between the charges, r = 0.15 m
The force of attraction between the charges will be calculated using Coulomb's law:
F = (k|q1q2|) / r^2
Where:
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 Nm^2/c^2
|q1| is magnitude of charge 1
|q2| is magnitude of charge 2
F = (9 x 10^9 x 7 x 10^-5 x 4 x 10^-5) / (0.15 x 0.15)
F = 1120 N
Thus, the force between the charges is 1120 N
In a collision between two unequal masses, how does the impulse imparted to the smaller mass by the larger mass compare with the impulse imparted to the larger mass by the smaller one?
Answer:
They are equal
Explanation:
Newton 3rd Law of motion states that for every force applied or action there is usually an equal and opposite force.
Impulse = Force * time
It is measured in Newton seconds.
The force and time of collision is the same which translates to an equal impulse by both scenarios.
The impulse imparted to the smaller mass by the larger mass is equal in magnitude to the impulse imparted to the larger mass by the smaller one, due to the conservation of momentum and Newton's third law. Both experience equal and opposite momentum transfers, ensuring the total momentum of the system remains constant.
Explanation:In the context of collisions, impulse is defined as the change in momentum of an object when it is subjected to a force over a period of time. According to Newton's third law, 'For every action, there is an equal and opposite reaction,' meaning that the impulse imparted to the smaller mass by the larger mass is exactly equal in magnitude to the impulse imparted to the larger mass by the smaller one, although the direction of the impulses will be opposite. When considering the conservation of momentum, the total momentum before the collision must equal the total momentum after the collision if no external forces are acting on the system (assuming a closed system). Therefore, if two cars collide, such as described in the provided text, regardless of their masses, the momentum transfer will be the same for both, thus the total momentum of the system remains constant.