Answer:
a) After 26.67 seconds it returns back to the origin
b) Velocity when it returns to the origin = 60 m/s in the -x direction
Explanation:
a) Let the starting position be origin and time be t.
After time t displacement, s = 0 m
Initial velocity, u = 60 m/s
Acceleration, a = -4.5 m/s²
We have equation of motion s = ut + 0.5 at²
Substituting
s = ut + 0.5 at²
0 = 60 x t + 0.5 x (-4.5) x t²
2.25t² - 60 t = 0
t² - 26.67 t = 0
t (t-26.67) = 0
t = 0s or t = 26.67 s
So after 26.67 seconds it returns back to the origin
b) We have equation of motion v = u + at
Initial velocity, u = 60 m/s
Acceleration, a = -4.5 m/s²
Time , t = 26.67
Substituting
v = 60 - 4.5 x 26.67 = -60 m/s
Velocity when it returns to the origin = 60 m/s in the -x direction
A 0.5-kg uniform meter stick is suspended by a single string at the 30-cm mark. A 0.2-kg mass hangs at the 80 cm mark. What mass hung at the 10-cm mark will produce equilibrium?
To find the mass needed to produce equilibrium, we perform a torque balance about the pivot point. This involves adding up the torques on each side of the pivot point and setting them equal to each other, then solving for the unknown mass at 10 cm.
Explanation:In this question, we are dealing with torque and equilibrium. For a system to be in equilibrium, the sum of the torques about any pivot point must be zero. Given a 0.5-kg uniform meter stick suspended by a single string at the 30-cm mark, with a 0.2-kg mass hanging at the 80-cm mark, we want to find the mass that should be hung at the 10-cm mark to ensure equilibrium.
To find this, we set up the following equation for the torques about the pivot point (the 30-cm mark):
Torque_due_to_0.5kg_rod + Torque_due_to_hanging_mass_at_10_cm = Torque_due_to_0.2kg_mass_at_80cm
Each torque can be calculated as the product of the force (which is the weight, or mass times gravity) and the distance from the pivot. After plugging in the given values and solving for the unknown mass at 10 cm, you will find the mass that should be hung at the 10 cm mark to produce equilibrium.
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If T is the absolute temperature, the intensity of radiation from an ideal radiator will be proportional to (a) T (b) T (c) T (d) T (e) none of the above
Answer:
T^4
Explanation:
According to the Stefan's law, the energy radiated per unit time and per unit area is directly proportional to the four power of teh absolute temperature of the black body.
E ∝ T^4
The intensity of radiation from an ideal radiator is not directly proportional to the absolute temperature T, but rather to T4, according to the Stefan-Boltzmann law of radiation.
Explanation:The intensity of radiation from an ideal radiator, or blackbody, is not directly proportional to the absolute temperature T; instead, it follows the Stefan-Boltzmann law of radiation. This law states that the intensity of radiation, referred to as 'P' or power, is directly proportional to the fourth power of the absolute temperature (P = σAT4). Here, σ (sigma) represents Stefan-Boltzmann constant, A denotes surface area, and T signifies absolute temperature.
This means the answer to your question is none of the above, as the intensity of the radiation isn't directly proportional to T, but rather to T4.
As an example, considering two temperatures T₁ = 293 K and T₂ = 313 K, the rate of heat transfer with T₂ is about 30 percent higher than T₁. This showcases the considerable impact of temperature on radiation intensity.
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The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2615 N with an effective perpendicular lever arm of 2.85 cm , producing an angular acceleration of the forearm of 110.0 rad / s2 . What is the moment of inertia of the boxer's forearm?
Answer:
0.68 kg-m²
Explanation:
F = Force applied by the muscle = 2615 N
r = effective perpendicular lever arm = 2.85 cm = 0.0285 m
α = Angular acceleration of the forearm = 110.0 rad/s²
I = moment of inertia of the boxer's forearm = ?
Torque is given as
τ = I α eq-1
Torque is also given as
τ = r F eq-2
using eq-1 and eq-2
r F = I α
(0.0285)(2615) = (110.0) I
I = 0.68 kg-m²
When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt? (a) How much torque are you exerting in newtons x meters (relative to the center of the bolt) (b) Convert this torque to footpounds
Answer:
Part a)
[tex]\tau = 23.1 Nm[/tex]
Part b)
[tex]\tau = 17.05 Foot pound force[/tex]
Explanation:
As we know that torque is defined as the product of force and its perpendicular distance from reference point
so here we have
[tex]\tau = \vec r \times \vec F[/tex]
now we have
[tex]\tau = (0.140)(165)[/tex]
[tex]\tau = 23.1 Nm[/tex]
Part b)
Now we know the conversion as
1 meter = 3.28 foot
1 N = 0.225 Lb force
now we have
[tex]\tau = 23.1 Nm[/tex]
[tex]\tau = 23.1 (0.225 Lb)(3.28 foot)[/tex]
[tex]\tau = 17.05 Foot pound force[/tex]
The torque exerted when tightening a bolt with a force of 165 N at a distance of 0.140 m from the center of the bolt is 23.1 N·m. This torque can be converted to approximately 17.0 ft·lb.
Explanation:To find the torque exerted when tightening a bolt, you can use the equation T = RF sin(θ). In this case, the force is 165 N and the distance from the center of the bolt is 0.140 m. Since you are pushing perpendicularly, the angle θ is 90 degrees. Thus, the torque exerted is 165 N x 0.140 m x sin(90°) = 23.1 N·m.
To convert this torque to foot-pounds, you can use the conversion factor 1 N·m = 0.73756 ft·lb. Therefore, the torque is 23.1 N·m x 0.73756 ft·lb/N·m ≈ 17.0 ft·lb.
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A car of mass 2800 kg collides with a truck of mass 4000 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car's velocity just before the collision was <36, 0, 0> m/s, and the truck's velocity just before the collision was <-15, 0, 29> m/s. (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. (d) What is the increase in internal energy of the car and truck (thermal energy and deformation)?
(a) The velocity of the car-truck system after collision is 18.08 m/s.
(b) The increase in internal energy of the car and truck is 2,835,031.2 J.
Velocity of the truck - car system after collision
The velocity of the system after collision is determined by applying the principle of conservation of linear momentum as shown below;
Final velocity in x - directionm₁u₁ + m₂u₂ = v(m₁ + m₂)
2800(36) + 4000(-15) = vx(2800 + 4000)
40,800 = 6800vx
vx = 6 m/s
Final velocity in z - directionm₁u₁ + m₂u₂ = v(m₁ + m₂)
2800(0) + 4000(29) = vz(2800 + 4000)
116,000 = 6800vz
vz = 17.06 m/s
Resultant velocity of the car-truck system after the collision [tex]v= \sqrt{v_x^2 + v_z^2} \\\\v = \sqrt{6^2 + 17.06^2} \\\\v = 18.08 \ m/s[/tex]Initial kinetic energy of the car and truckK.E(car) = ¹/₂mv²
K.E(car) = ¹/₂ x (2800) x (36)²
K.E(car) = 1,814,400 J
v(truck) = √(15² + 29²) = 32.65
K.E(truck) = ¹/₂ x (4000) x (32.65)²
K.E(truck) = 2,132,045 J
K.E(total) = 1,814,400 J + 2,132,045 J = 3,946,445 J
Final kinetic energy of the systemK.E = ¹/₂(m₁ + m₂)v²
K.E = ¹/₂ x (2800 + 4000) x (18.08)²
K.E = 1,111,413.8 J
Increase in internal energyU = ΔK.E
U = 3,946,445 J - 1,111,413.8 J
U = 2,835,031.2 J
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Final answer:
To determine the velocity of the stuck-together car and truck after the collision, apply the principle of conservation of momentum. Consider the change in kinetic energy and the production of thermal energy and deformation.
Explanation:
To determine the velocity of the stuck-together car and truck just after the collision, we can apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. We can find the final velocity by summing the momenta of the car and truck and dividing by their combined mass.
In this case, the car's momentum is the product of its mass and velocity, and the truck's momentum is the product of its mass and velocity. Adding these momenta together and dividing by the combined mass gives us the final velocity of the stuck-together car and truck just after the collision.
For the increase in internal energy of the car and truck, we need to consider the change in kinetic energy and the production of thermal energy and deformation. The change in kinetic energy can be calculated by finding the difference between the initial and final kinetic energies of the car and truck. The thermal energy and deformation depend on factors such as the materials involved and the severity of the collision.
A rigid tank contains 1.80 moles of helium, which can be treated as an ideal gas, at a pressure of 27.3 atm. While the tank and gas maintain a constant volume and temperature, a number of moles are removed from the tank, reducing the pressure to 5.20 atm. How many moles are removed?
Answer:
Has been removed 1.458 moles.
Explanation:
n1= 1.8 mol
p1= 27.3 atm
p2= 5.2 atm
n2= ?
n2= n1 * p2/p1
n2= 0.342 moles
Δn= n1-n2
Δn= 1.458 moles
The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary to increase the object's temperature from 6°C to 12°C? 4Q 2Q Q 3Q
Answer:
The answer is 3Q
Explanation:
The metal temperature increases in a linear way, we could get a difference between final and initial temperature
[tex]DT=FinalTemperature-InitialTemperature[/tex]
We get a temperature difference of 2 degrees per each heat addition. If we add the same heat 3 times more, it will increase to 12 degrees
Answer:
The quantity of heat required to increase the temperature of the object from 6°C to 12°C is 3Q
Explanation:
Heat capacity is the quantity of heat required to increase the temperature of an object.
Q = mcΔθ
where;
Q is the quantity of heat
m is the mass of the object
c is specific heat capacity of the object
Δθ is change in temperature = T₂ - T₁
For the first sentence of this question;
Q = mc(6-4)
Q = mc(2)
Q = 2mc
For the second sentence of this question;
Let Q₂ be the quantity of heat required to increase the temperature of the object from 6°C to 12°C
Q₂ = mcΔθ
Q₂ = mc(12-6)
Q₂ = mc(6)
Q₂ = 6mc
Q₂ = 3(2mc)
Recall, Q = 2mc
Thus, Q₂ = 3Q
A 0.7 kg lab cart moving to the right at 0.15 m/s collides with a 0.5 kg lab cart moving to the right at 0.10 m/s. After the collision, the 0.7 kg cart is moving to the right at 0.08 m/s. Calculate the velocity of the 0.5 kg lab cart after the collision.
Answer:
0.198 m/s
Explanation:
m1 = 0.7 kg, u1 = 0.15 m/s,
v1 = 0.08 m/s
m2 = 0.5 kg, u2 = 0.1 m/s
Let the speed of 0.5 kg is v2 after the collision.
By using the conservation of momentum
Momentum before collision = momentum after collision
m1 u1 + m2 u2 = m1 v1 + m2 v2
0.7 x 0.15 + 0.5 x 0.1
= 0.7 × 0.08 + 0.5 × v2
0.099 = 0.5 v2
v2 = 0.198 m/s
What is the difference between the formation of an ionic bond and formation of a covalent bond?
Answer:
Answer to the question:
Explanation:
Differences between ionic bond and covalent bond:
The ionic bond occurs between two different atoms (metallic and non-metallic), while the covalent bond occurs between two equal atoms (non-metallic).
In the covalent bond there is an electron compartment, while in the ionic bond there is an electron transfer.
Ionic bonds have a high melting and boiling point, while covalent bonds usually have a low point.
Final answer:
An ionic bond involves the transfer of electrons, resulting in a bond between a positive ion and a negative ion. A covalent bond involves the sharing of electrons between atoms, creating a stable electron arrangement for both atoms.
Explanation:
An ionic bond is formed by the transfer of electrons from one atom to another, resulting in a bond between a positive ion (cation) and a negative ion (anion). On the other hand, a covalent bond is formed by the sharing of electrons between atoms, creating a bond in which electrons are shared rather than transferred.
In an ionic bond, one atom gains electrons and becomes negatively charged, while the other atom loses electrons and becomes positively charged. This attraction between oppositely charged ions forms the bond. In a covalent bond, atoms share one or more pairs of electrons, resulting in a stable electron arrangement for both atoms.
Examples of compounds with ionic bonds include sodium chloride (NaCl) and magnesium oxide (MgO), while examples of compounds with covalent bonds include water (H2O) and methane (CH4).
Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilometers will the train have traveled after 4.00 hours? Round to the nearest km.
Answer:
605 km
Explanation:
Hello
the same units of measure should be used, then
Step 1
convert 42 m/s ⇒ km/h
1 km =1000 m
1 h = 36000 sec
[tex]42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\[/tex]
[tex]0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}[/tex]
Step 2
find kilometers traveled after 4 hours
[tex]V=\frac{s}{t}\\ \\[/tex]
V,velocity
s, distance traveled
t. time
now, isolating s
[tex]V=\frac{s}{t} \\s=V * t\\[/tex]
and replacing
[tex]s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\[/tex]
S=604.8 Km
Have a great day
Gina perceives the car to be far away because the sides of the road upon which it is moving seem to come together to be no wider than the car itself. This is an example of the ____ cue to depth.
Answer:
Linear Perspective.
Explanation:
Linear perspective cue to depth and it is in respect to both texture gradient or the next depth cue and relative size. In linear perspective, parallel lines that go along seems to have converged at distance very far, eventually reaching a vanishing point at the horizon. For example the railway track appears to be converging at a point very distant from the observer but it didn't.
Answer:
d. linear perspective
Explanation:
According to a different source, these are the options that come with this question:
a. visual acuity
b. texture gradient
c. retinal disparity
d. linear perspective
Linear perspective refers to a way in which depth is created. This term is particularly employed by artists who create an illusion of depth on a flat surface. This perspective leads all parallel lines to be lost in a single vanishing point on the horizon line. In this example, Gina is using the linear perspective cue to understand the depth of the scene she is observing.
Sound 1 has an intensity of 45.0 W/m2. Sound 2 has an intensity level that is 3.2 dB greater than the intensity level of sound 1. What is the intensity level of sound 2?
Answer:
The intensity level of sound 2 is 93.3\ W/m².
Explanation:
Given that,
Intensity of sound 1 = 45.0 W/m²
Intensity level of sound 2 = 3.2 dB
[tex]I_{0}=10^{-12}\ w/m^2[/tex]
We need to calculate the intensity
Using equation of the sound level intensity
[tex]I_{dB}=10 log\dfrac{I}{I_{0}}[/tex]
[tex]I_{dB}=10 log(\dfrac{45.0}{10^{-12}})[/tex]
[tex]I_{dB}=136.5 dB[/tex]
The intensity of sound 2 is greater than 3.2 dB.
Therefore,
[tex]I_{2}=136.5+3.2[/tex]
[tex]I_{2}=139.7\ dB[/tex]
Calculate the intensity of sound 2
[tex]I_{2}=10log(\dfrac{I'}{10^{-12}})[/tex]
[tex]I'=10^{-12}(10^{\frac{139.7}{10}})[/tex]
[tex]I'=93.3\ W/m^2[/tex]
Hence, The intensity level of sound 2 is 93.3\ W/m².
Decibels, measured in dB, quantifies the intensity level of sound. They relate to the intensity of sound in a logarithmic manner. More information and advanced calculations are needed to find the actual intensity of Sound 2 based on the given decibel difference
Explanation:Let's begin by understanding that decibels (dB) measure the intensity level of sound. The decibel level (dB) is a logarithmic unit used to describe a ratio of two values of a physical quantity, here, the sound intensity. One key fact is that each time the intensity increases by a factor of 10; the decibel level rises by 10 dB. Thus, the difference in decibels corresponds to the logarithmic ratio of the two intensities involved.
In the given problem, Sound 2 is 3.2 dB greater than Sound 1. It means the ratio of their intensities has a logarithmic relation with a value of 3.2. However, to convert this decibel difference into an actual value for Sound 2's intensity, additional calculations involving logarithms and the initial intensity value for Sound 1 are needed. As these calculations involve advanced math, it'd be more appropriate to take this problem to a higher level course or consult a logarithm table or a scientific calculator.
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In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of the wind and the speed of the airplane in still air. g
Answer:
Velocity of airplane is 500 km/h
Velocity of wind is 40 km/h
Explanation:
[tex]V_a[/tex]= Velocity of airplane in still air
[tex]V_w[/tex]= Velocity of wind
Time taken by plane to travel 1150 km against the wind is 2.5 hours
[tex]V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)[/tex]
Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours
[tex]V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)[/tex]
Subtracting the two equations we get
[tex]V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h[/tex]
Applying the value of velocity of wind to the first equation
[tex]V_a-40=460\\\Rightarrow V_a =500\ km/h[/tex]
∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h
The given mathematical problem involves calculating speeds. The speed of the airplane in still air was found to be 501 km/h and the wind speed was calculated to be 41 km/h, by using the concept of speed equals distance divided by time.
Explanation:The subject is related to measurements of speed, which is essentially a mathematical problem, specifically involving algebraic calculations. In order to solve this problem, we will use the concept that speed equals distance divided by time.
Firstly, convert all time measurements to the same unit. Since the speed of an airplane is typically measured in km/h, it will be convenient to convert the 50 minutes to hours (50/60 = 0.83 hours).
When moving against the wind, the combined speed (airplane speed minus wind speed) is 1150 km / 2.5 h = 460 km/h. When moving with the wind, the combined speed (airplane speed plus wind speed) becomes 450 km / 0.83 h = 542 km/h.
Now you can find the speed of the airplane in still air by averaging the two combined speeds: (460 km/h + 542 km/h) / 2 = 501 km/h. The speed of the wind can be found by subtracting the airplane's speed from the higher combined speed: 542 km/h - 501 km/h = 41 km/h. Therefore, the speed of the airplane in still air is 501 km/h and the wind speed is 41 km/h.
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Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s. What temperature does this represent?
Answer:
[tex]T = 1.26 \times 10^8 K[/tex]
Explanation:
As we know that rms speed of ideal gas is given by the formula
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
here we know that
[tex]v_{rms} = 2.67 \times 10^5 m/s[/tex]
molecular mass of gas is given as
[tex]M = 44 g/mol = 0.044 kg/mol[/tex]
now from above formula we have
[tex]2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}[/tex]
now we have
[tex]T = 1.26 \times 10^8 K[/tex]
The root mean square speed of a molecule in an ideal gas can be used to determine the temperature of the gas when vrms is known. This involves solving the formula for temperature using the given vrms and mass of a molecule. We need to ensure that the molar mass is properly converted to the mass of a molecule and that we use SI units.
Explanation:The subject question is related to the concept of root mean square speed (vrms) in gas physics and involves calculations using the ideal gas model. According to this concept, we know that vrms is given by the equation vrms = sqrt(3kB T / m), where:
kB is the Boltzmann constant (1.38×10−23 J/K), T is the temperature in Kelvin, and m is the mass of each (identical) particle in the gas.Given that vrms = 2.67 x 105 m/s and molar mass (M) = 44.0 g/mol (which must be converted to kg as SI units are used in this context), we can solve the formula for T to find the temperature T = vrms^2×m / (3×kB) after substituting the known values into this equation. Note that one must convert the molar mass to mass of a molecule by dividing it by Avogadro's number (6.022 x 10²³ mol⁻¹).
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You are in a fire truck heading directly toward a tunnel in a mountainside at 17m/s. you hear the truck's 90Hz siren as well as its reflection from the mountainside. What beat frequency do you hear?
Hey there!:
Take the speed of sound to be 343m/s.
Direct frequency perceived by observer:
(343 + 17) / 343) * 90Hz = 94.460Hz
Change in frequency = ( 4.460 - 90 ) = 4.460Hz.
90 - (4.460 x 2) = 81.08 Hz. indirect frequency heard by observer.
Therefore :
Beat frequency = (94.460 - 81.08) = 13.38Hz.
Hope this helps!
A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?
Answer:
3.95 m
Explanation:
m = 1 kg, h = 100 m, k = 125 N/m
Let the spring is compressed by y.
Use the conservation of energy
potential energy of the mass is equal to the energy stored in the spring
m x g x h = 1/2 x ky^2
1 x 9.8 x 100 = 0.5 x 125 x y^2
y^2 = 15.68
y = 3.95 m
A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30°. The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the magnitude of the frictional force on the sphere?
Answer:
40N
Explanation:
frictional force = mgsin©
frictional force = 8*10*sin30
frictional force = 8*10*0.5= 40N//
The emf of a battery is equal to its terminal potential difference: A) under all conditions B) only when the battery is being charged C) only when a large cwrrent is in the battery D) only when there is no current in the battery E) under no conditions
Answer:
option (C)
Explanation:
EMF stand for electro motive force. the emf of a battery is the potential between the two electrodes when it is not use in the circuit.
The terminal potential difference is the potential difference between the electrodes of a cell when it is in use.
EMF is only when the current is very large in the battery.
The emf of a battery is equal to its terminal potential difference only when a large cwrrent is in the battery.
C) only when a large cwrrent is in the battery.
A vertically mounted spring (k = 750 N/m) is compressed by 35.0 cm relative to its unstrained length. A mass (m = 0.36 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. How high from release point the mass can reach?
Answer:
Height, h = 13.02 meters
Explanation:
It is given that,
Spring constant of the spring, k = 750 N/m
It is compressed by 35 cm relative to its unstained length, x = 35 cm = 0.35 m
Mass of the object, m = 0.36 kg
When the spring is released, the mass is launched vertically in the air. We need to find the height attained by the mass at this position. On applying the conservation of energy as :
Energy stored in the spring = change on potential energy
[tex]\dfrac{1}{2}kx^2=mgh[/tex]
[tex]h=\dfrac{kx^2}{2mg}[/tex]
[tex]h=\dfrac{750\ N/m\times (0.35\ m)^2}{2\times 0.36\ kg\times 9.8\ m/s^2}[/tex]
h = 13.02 meters
So, the mass will reach a height of 13.02 meters. Hence, this is the required solution.
The maximum height reached by the mass when the spring is released is 13.02 m.
The given parameters;
spring constant, k = 750 N/mextension of the spring, x = 35 cm = 0.35 mmass attached, m = 0.36 kgApply the principle of conservation energy to determine the maximum height reached by the mass when the spring is released.
mgh = ¹/₂kx²
where;
h is the maximum height reached by the massg is acceleration due to gravity[tex]h = \frac{kx^2}{2mg} \\\\h = \frac{(750) \times (0.35)^2}{2(0.36\times 9.8)} \\\\h = 13.02 \ m[/tex]
Thus, the maximum height reached by the mass when the spring is released is 13.02 m.
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An object with mass 80 kg moved in outer space. When it was at location <7, -34, -7> its speed was 14.0 m/s. A single constant force <200, 460, -150> N acted on the object while the object moved to location <12, -42, -11> m. What is the speed of the object at this final location?
The speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.
What is the speed of a body?The speed of a body is the rate at which it covers the total distance is in the time taken.
If the speed of the body has direction, then it is known as the velocity. The velocity is the vector quantity.
The constant force applied on the object is <200, 460, -150> N. It can be written in the vector form as,
[tex]\vec F=200\hat i+460\hat j-150\hat k[/tex]
The initial position vector of the object for the location <7, -34, -7> can be given as,
[tex]d_i=7\hat i-34\hat j-7\hat k[/tex]
The final position vector of the object is for the location <12, -42, -11> m can be given as,
[tex]d_f=12\hat i-42\hat j-11\hat k[/tex]
Thus, the total distance traveled by the constant force is,
[tex]d=(12\hat i-42\hat j-11\hat k)-(7\hat i-34\hat j-7\hat k)\\d=5\hat i-8\hat j-4\hat k[/tex]
The work done by a object is the product of force and displacement. Thus, work done on the object is,
[tex]W=\vec F.\vec d \\W=(200\hat i+460\hat j-150\hat k).(5\hat i-8\hat j-4\hat k)\\W=1000-3680+600\\W=-2080 \rm J[/tex]
Now, the work done on a object is equal to the kinetic energy of the body. Thus, work done on the object is,
[tex]W=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]
Here, v(f) is the final speed and v(i) is the initial speed of the object. As the mass of the object is 80 kg and initial speed is 14 m/s. Thus, put the values as,
[tex]-2080=\dfrac{1}{2}(80)(v^2_f-14^2)\\v_f=12\rm m/s[/tex]
Thus, the speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.
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To find the speed of the object at the final location, we need to calculate the net force acting on it using the equation F = ma. By plugging in the values, we can calculate the final speed of the object.
Explanation:To determine the speed of the object at the final location, we need to calculate the net force acting on the object using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.
Given that the force acting on the object is <200, 460, -150> N, and the mass of the object is 80 kg, we can calculate the acceleration using the formula a = F/m. After calculating the acceleration, we can use it to find the final speed of the object using the equation v^2 = u^2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the displacement.
By plugging in the given values, we can calculate the final speed of the object at the location <12, -42, -11> m.
A good baseball pitcher can throw a baseball toward home plate at 87 mi/h with a spin of 1710 rev/min. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 ft path is a straight line.
Explanation:
First, convert 87 mi/h to ft/min.
87 mi/h × (5280 ft/mi) × (1 h / 60 min) = 7656 ft/min
The time to reach the home plate is:
t = 60 ft / 7656 ft/min
t = 0.00784 min
The number of revolutions made in that time is:
n = 1710 rev/min × 0.00784 min
n = 13.4 rev
Rounding to 2 significant figures, the ball makes 13 revolutions.
You make a capacitor from 2 flat plates each with an area of 10 cm^2 you use a 1mm thick piece of pyrex glass as your dielectric. a. a.what is the maximum voltage you can safely apply to this capacitor?
b. what is its capacitance?
c. what is the maximum amount of charge you can store?
d. when this much charge is on the capacitor, how much energy is stored?
Answer:
Part a)
[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]
Part b)
[tex]C = 4.96 \times 10^{-11} farad[/tex]
Part c)
[tex]Q = 0.69 \mu C[/tex]
Part d)
[tex]E = 4.86 \times 10^{-3} J[/tex]
Explanation:
Part a)
As we know that dielectric constant of pyrex glass is 5.6 and its dielectric breakdown strength is given as
[tex]E = 14 \times 10^6 V/m[/tex]
now we have
[tex]E . d = \Delta V[/tex]
[tex](14 \times 10^6)(0.001) = \Delta V[/tex]
so we have
[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]
Part b)
Capacitance is given as
[tex]C = \frac{k\epsilon_0 A}{d}[/tex]
[tex]C = \frac{5.6(8.85 \times 10^{-12})(10 \times 10^{-4}}{0.001}[/tex]
[tex]C = 4.96 \times 10^{-11} farad[/tex]
Part c)
Now we have
[tex]Q = C\Delta V[/tex]
[tex]Q = (4.96 \times 10^{-11})(14 \times 10^3)[/tex]
[tex]Q = 0.69 \mu C[/tex]
Part d)
[tex]Energy = \frac{1}{2}CV^2[/tex]
[tex]E = \frac{1}{2}(4.96 \times 10^{-11})(14 \times 10^3)^2[/tex]
[tex]E = 4.86 \times 10^{-3} J[/tex]
The blackbody radation emmitted from a furnace peaks at a wavelength of 1.9 x 10^-6 m (0.0000019 m). what is the temperature inside the furnace?
Answer:
Temperature, T = 1542.10 K
Explanation:
It is given that,
The black body radiation emitted from a furnace peaks at a wavelength of, [tex]\lambda=1.9\times 10^{-6}\ m[/tex]
We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.
[tex]\lambda\propto \dfrac{1}{T}[/tex]
or
[tex]\lambda=\dfrac{b}{T}[/tex]
b = Wein's displacement constant
[tex]\lambda=\dfrac{2.93\times 10^{-3}}{T}[/tex]
[tex]T=\dfrac{2.93\times 10^{-3}}{\lambda}[/tex]
[tex]T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}[/tex]
T = 1542.10 K
So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.
Using Wien's Displacement Law, the temperature inside a furnace emitting radiation that peaks at a wavelength of 1.9 x 10^-6 m is approximately 1525 Kelvin.
Explanation:The temperature of the furnace can be determined using Wien's Displacement Law, which states that the product of the temperature of a black body and the wavelength at which the radiation it emits is most intense is approximately equal to 2.898 x 10^-3 m.K. In this specific case, the peak wavelength of the energy emitted from the furnace is given as 1.9 x 10^-6 m. Therefore, the temperature inside the furnace can be calculated using the formula: T = Amax / λ, where Amax = 2.898 x 10^-3 m.K and λ = 1.9 x 10^-6 m. Calculating from the given formula, T ≈ 1525 K. Therefore, the temperature inside the furnace is approximately 1525 Kelvin.
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A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the position function? (c) When is the velocity zero?
Explanation:
It is given that,
A particle starts from rest and has an acceleration function as :
[tex]a(t)=(5-10t)\ m/s^2[/tex]
(a) Since, [tex]a=\dfrac{dv}{dt}[/tex]
v = velocity
[tex]dv=a.dt[/tex]
[tex]v=\int(a.dt)[/tex]
[tex]v=\int(5-10t)(dt)[/tex]
[tex]v=5t-\dfrac{10t^2}{2}=5t-5t^2[/tex]
(b) [tex]v=\dfrac{dx}{dt}[/tex]
x = position
[tex]x=\int v.dt[/tex]
[tex]x=\int (5t-5t^2)dt[/tex]
[tex]x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3[/tex]
(c) Velocity function is given by :
[tex]v=5t-5t^2[/tex]
[tex]5t-5t^2=0[/tex]
t = 1 seconds
So, at t = 1 second the velocity of the particle is zero.
To find the velocity function of the particle, integrate the acceleration function, which results in 5t - 5t^2. The position function is found by integrating the velocity function, yielding 1/2 * 5t^2 - 1/3 5t^3. The velocity is zero at t = 0 and t = 1 second.
We can solve for the velocity function and the position function by integrating the acceleration function.
Velocity Function
The velocity function is the integral of the acceleration function. Since the acceleration function given is a(t) = 5 - 10t m/s2, the velocity function v(t) is obtained by integrating a(t) with respect to time:
v(t) = ∫ (5 - 10t) dt = 5t - 5t2 + C
Since the particle starts from rest, the initial velocity is 0, which means C = 0. Therefore, the velocity function is v(t) = 5t - 5t2.
Position Function
The position function s(t) is the integral of the velocity function. We integrate v(t) to obtain the position function:
s(t) = ∫ (5t - 5t2) dt = ½ 5t2 - ⅓ 5t3 + K
Since the particle starts from position zero, K = 0 and the position function is s(t) = ½ 5t2 - ⅓ 5t3.
Zero Velocity
To find when the velocity is zero, we set the velocity function equal to zero and solve for t:
5t - 5t2 = 0t = 0 or t = 1 s.
Work is required to compress 5.00 mol of air at 20.00C and 1.00 atm to one-tenth of the original volume by an adiabatic process. How much work is required to produce this same compression?
Answer:46.03 KJ
Explanation:
no of moles(n)=5
temperature of air(T)=[tex]20^{\circ}[/tex]
pressure(p)=1 atm
final volume is [tex]\frac{V}{10}[/tex]
We know work done in adaibatic process is given by
W=[tex]\frac{P_iV_i-P_fV_f}{\gamma -1}[/tex]
[tex]\gamma[/tex] for air is 1.4
we know [tex]P_iV_i^{\gamma }= P_fV_f^{\gamma }[/tex]
[tex]1\left ( V\right )^{\gamma }[/tex]=[tex]P_f\left (\frac{v}{10}\right )^{\gamma }[/tex]
[tex]10^{\gamma }=P_f[/tex]
[tex]P_f=25.118 atm[/tex]
W=[tex]\frac{1\times 0.1218-25.118\times 0.01218}{1.4 -1}[/tex]
W=-46.03 KJ
it means work is done on the system
Two banked curves have the same radius. Curve A is banked at 12.7 °, and curve B is banked at an angle of 15.1°. A car can travel around curve A without relying on friction at a speed of 19.1 m/s. At what speed can this car travel around curve B without relying on friction?
The speed of the car on Curve B is obtained by solving for the radius in Curve A's equation, then substituting that into the formula for Curve B. The formula involved is based on principles of circular motion and forces.
Explanation:The subject of this question is Physics, specifically the principles of circular motion and the forces at play within. Given the information from Curve A, we can ascertain that the speed of the car on Curve B can be found using principles of physics. The formula to find the speed at which the car can travel around a banked curve without relying on friction is: v = sqrt[rgtan(Θ)]. Where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and Θ is the angle of the banked curve.
Since the radius is the same for both curves, and the speed is known for Curve A, we can set it up so: 19.1 = sqrt[r * 9.8 * tan(12.7)]. We then solve for r (the radius), and apply it to Curve B: v = sqrt[r * 9.8 * tan(15.1)]. By substituting in the value of r obtained from the first equation, we can calculate the speed for Curve B.
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To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.
Explanation:The speed at which a car can travel around a banked curve without relying on friction can be calculated using the ideal banking angle formula. The formula is given by v = √(g * r * tan(θ)), where v is the speed, g is the acceleration due to gravity, r is the radius of the curve, and θ is the banked angle. In this case, curve A is banked at 12.7° and the car can travel at a speed of 19.1 m/s. To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.
Plugging in the values into the formula, v = √(9.8 * r * tan(15.1)). Since the radius is the same for both curves, we can solve for v: 19.1 = √(9.8 * r * tan(12.7))
Solving the equation for v, we find that the car can travel around curve B without relying on friction at a speed of approximately 21.2 m/s.
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A 1 cm^3 block with a density of 0.92 g/cm^3 is floating in a container of water (d =1g/ cm^3). You may ignore any air pressure throughout this problem. What buoyant force is necessary to keep the block from sinking?
Answer:
980 dyne
Explanation:
Volume = 1 cm^3, d = 0.92 g / cm^3, D = 1 g/cm^3
In the equilibrium condition, the buoyant force is equal to the weight of the block.
Buoyant force = Volume of block x density of water x g
Buoyant force = 1 x 1 x 980 = 980 dyne
A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
Answer:
a) Maximum height reached above ground = 2.8 m
b) When he reaches maximum height he is 2 m far from end of the ramp.
Explanation:
a) We have equation of motion v²=u²+2as
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0²=5.6² + 2 x -9.81 x s
s = 1.60 m
Height above ground = 1.2 + 1.6 = 2.8 m
b) We have equation of motion v= u+at
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0= 5.6 - 9.81 x t
t = 0.57s
Now considering horizontal motion of skateboarder.
We have equation of motion s =ut + 0.5 at²
u = 6.6cos58 = 3.50 m/s
a = 0 m/s²
t = 0.57
Substituting
s =3.5 x 0.57 + 0.5 x 0 x 0.57²
s = 2 m
When he reaches maximum height he is 2 m far from end of the ramp.
The highest point reached by the skateboarder is 1.612 meters above the ground. When the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is 3.568 meters.
Given:
Initial velocity (v₀) = 6.6 m/s
Launch angle (θ) = 58°
Height of the ramp (h) = 1.2 m
Acceleration due to gravity (g) = 9.8 m/s²
(a) To find the maximum height reached by the skateboarder:
Δy = v₀y² / (2g)
v₀y = v₀ × sin(θ)
v₀y = 6.6 × sin(58°)
v₀y = 5.643 m/s
Δy = (5.643 )² / (2 × 9.8)
Δy ≈ 1.612 m
Therefore, the highest point reached by the skateboarder is 1.612 meters above the ground.
(b) The time of flight can be calculated using the equation:
t = 2 × v₀y / g
t = 2 × 5.643 / 9.8
t = 1.153 s
Δx = v₀x × t
First, we need to find the initial horizontal velocity (v₀x):
v₀x = v₀ × cos(θ)
v₀x = 6.6 × cos(58°)
v₀x = 3.099 m/s
Δx = 3.099 * 1.153
Δx = 3.568 m
Therefore, when the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is 3.568 meters.
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Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to Lonnie with a return speed of 60 m/s. If the bat is in contact with the ball for 0.050 s what is the impulse experienced by the ball? A. 360 N.s B. 20 N.s C. 400 N.s D. 9.0 N.s
Answer:
B) 20N.s is the correct answer
Explanation:
The formula for the impulse is given as:
Impulse = change in momentum
Impulse = mass × change in speed
Impulse = m × ΔV
Given:
initial speed = 40m/s
Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)
mass = 0.20 kg
Thus, we have
Impulse = 0.20 × (40m/s - (-60)m/s)
Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s
The impulse experienced by the ball is,
[tex]\rm Impulse = 20\; Nsec[/tex]
Given :
Mass = 0.20 Kg
Initial Speed = 40 m/sec
Final Speed = -60 m/sec (minus sign shows that ball move in opposite direction)
Solution :
We know that the formula of Impulse is,
[tex]\rm Impulse = m\times \Delta V[/tex]
where, m is the mass and [tex]\rm \Delta V[/tex] is change in velocity.
[tex]\rm Impulse = 0.20\times(40-(-60))[/tex]
[tex]\rm Impulse = 20\; Nsec[/tex]
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A 50-kg crate is initially at rest on a horizontal floor. A person pushes horizontally with a constant force of 250 N. The maximum static frictional force between the floor and the crate is 275 N. What is the frictional force on the crate while the person pushes on it?
Answer:
250 N
Explanation:
m = Mass of cart = 50 kg
F = Pushing Force = 250 N
g = Acceleration due to gravity = 9.81 m/s²
μ = Coefficient of friction
[tex]F_s =\text{Static frictional force between the floor and the crate}= 275\ N[/tex]
N = Weight of cart = mg
N = 50×9.81
N = 490.5 N
[tex]F_s =\mu N\\\Rightarrow \mu=\frac{F_s}{N}\\\Rightarrow \mu=\frac{275}{490.5}\\\Rightarrow \mu=0.56[/tex]
Here it can be seen that the pushing force is less than the static frictional force so the crate will not move
∴ The frictional force on the crate while the person pushes on it is 250 N