ω = ?
mass = 45.8kg
r = 0.512m
E = 2.99*10⁹J
Kinetic Energy of rotation = I * ω²
K.E = I * ω²
I = ½ m*r²
I = ½ * 45.8 * (0.512)²
I = 6.0kgm²
K.E = ½ * I * ω²
ω = √(2K.E / I )
ω = √[( 2* 2.99*10⁹) / 6]
ω = 3.157*10⁴ rad/s
If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive charge and the same amount of negative charge. If you placed these charges 8.00 m apart, how strong would the attractive forces between them be?
Answer:
The attractive force between them is [tex]1.296 \times 10^{18}[/tex] N
Explanation:
Given:
Charge [tex]q = 96000[/tex] C
Distance between two charges [tex]r = 8[/tex] m
According to the coulomb's law,
[tex]F = \frac{kq^{2} }{r^{2} }[/tex]
Where [tex]k = 9 \times 10^{9}[/tex] = force constant.
[tex]F = \frac{9 \times 10^{9} \times (96000)^{2} }{8^{2} }[/tex]
[tex]F = 1.296 \times 10^{18}[/tex] N
Therefore, the attractive force between them is [tex]1.296 \times 10^{18}[/tex] N
Draw equipotential lines near the positive and negative charges below with dashed lines. b) Draw solid electric field lines based on the fact that they are perpendicular to the equipotential lines at every point. Include arrows on the electric field lines to indicate their direction (since they represent vectors at each location).
Answer: find the attached figure for a and b
Explanation:
A) The second figure depict electric field lines and equipotential lines for two equal but opposite charges. The equipotential lines can be drawn by making them perpendicular to the electric field lines. The potential is greatest (most positive) near the positive charge and least (most negative) near the negative charge.
B) The figure attached depicts an isolated point charge Q with its electric field lines in blue and equipotential lines in green. The potential is the same along each equipotential line, meaning that no work is required to move a charge anywhere along one of those lines. Work is needed to move a charge from one equipotential line to another. Equipotential lines are perpendicular to electric field lines in every case.
Please find the attached file for the figure
The five general principles from the APA are meant to __________. A. be enforceable rules B. be posted in every office C. guide and inspire good conduct D. stand up in court Please select the best answer from the choices provided A B C D
Answer: C
guide and inspire good conduct
Explanation:
APA are rules of proper source documentation to avert difficulty. This makes you feel that you are not alone. Having to follow detailed citation and formatting guidelines as well may seem like just one more task to add to an already-too-long list of requirements.
Final answer:
Correct answer is C. The five general principles of the APA are meant to guide and inspire good conduct among psychologists, offering a values framework rather than serving as enforceable rules.
Explanation:
The five general principles from the American Psychological Association (APA) are meant to guide and inspire good conduct. These principles serve as ethical guidelines for behaviors and decision-making in psychological practice.
In contrast to enforceable rules, which are mandatory and spell out specific prohibitions and mandates, principles and guidelines such as those provided by the APA are not meant to be enforceable or stand up in court but rather to provide a framework for ethical practice.
They offer a set of values that psychologists are encouraged to aspire to in their professional conduct. These principles cover important areas such as respect for people's rights, integrity, and social responsibility in practice.
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).Smallest to largesta.coasting universeb.critical universec.recollapsing universe
Answer:
From smallest ratio to the largest ratio:
Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)
Explanation:
The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1
The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1
Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1
From smallest to largest ratio of actual mass density to critical density, the models are: coasting universe, critical universe, and recollapsing universe. In a coasting universe, expansion continues at a decreasing rate. A critical universe's expansion slows to an eventual stop. A recollapsing universe's expansion ultimately reverses, leading to contraction.
Explanation:The long-term expansion and possible contraction of the universe under different scenarios or models, in the absence of dark energy, can be ranked based on the ratio of its actual mass density to the critical density.
Continuing from the smallest mass density ratio to the largest, it would be: the coasting universe, the critical universe, and finally the recollapsing universe.
In a coasting universe, the actual density is lower than the critical density. Thus, it would continue to expand forever but at a decreasing rate. In a critical universe, the actual density equals the critical density, which means the universe's expansion will gradually slow to a stop in an infinitely far future. Lastly, in a recollapsing universe, the actual density is greater than the critical density. This leads to an eventual reversal of the expansion, causing the universe to start contracting.
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An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kpa and the evaporator at -12 degrees
c. if the cooling load is 150 kw, determine the cop and the power required for this service in kw.
Answer:
cop = 4.859
power = 30.87 KW
Explanation:
the pictures attached herewith shows the calculation
By calculating the work input to the compressor (Wc) using the COP formula and known cooling load (Qc), and assuming a COP of 4, the power required is 37.5 kW.
To determine the Coefficient of Performance (COP) and power required for the ideal vapor-compression refrigeration cycle, we'll use the following steps and equations:
Find the Heat Transfer in the Evaporator (Qc):
Given Cooling Load (Qc) = 150 kW.
Calculate Work Input to the Compressor (Wc):
Using the first law of thermodynamics:
Qc = Wc + Qh
Rearrange to find Wc:
Wc = Qc - Qh
Determine Heat Rejected in the Condenser (Qh):
To find Qh, we need to calculate the change in enthalpy in the evaporator and condenser using refrigerant properties, such as specific enthalpy values. For refrigerant R-134a at -12 degrees Celsius, you can use property tables or software to find the enthalpy values.
Find COP:
COP is defined as:
COP = Qc / Wc
Calculate Power Required for Service:
Using the COP and Qc:
Wc = Qc / COP
Now, let's calculate:
Assuming a COP value (e.g., 4):
Wc = 150 kW / 4 = 37.5 kW
So, with a chosen COP of 4, the power required for the service is 37.5 kW.
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Nuclear power stations transfer useful energy to other energy stores. How is this useful energy carried from the power stations?
I
Nuclear power stations transfer energy by nuclear fission to heat water, producing steam. This steam drives turbines connected to generators, producing electricity. The electricity is transported through power lines across the national grid.
Explanation:In a nuclear power station, the useful energy is primarily carried from the power plant through the national grid system. Here's how it works: nuclear power stations transfer energy via a process called nuclear fission where atoms are split, releasing a large amount of energy. This energy is used to heat up water, even to the point of converting it to steam. The steam drives the turbines which are connected to generators. These generators produce electricity, which is then distributed through power lines across the national grid, reaching homes, factories, offices and so on.
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Order the electromagnetic waves from lowest
frequency (1) to highest (4).
Gamma:
Infrared:
X-ray:
Visible light:
The electromagnetic waves from the lowest frequency to the highest frequency are gamma rays, X-ray, visible light, and infrared.
What are electromagnetic waves?Electromagnetic waves are the waves which are formed when an electric field couples with a magnetic field. Magnetic and the electric fields of an electromagnetic wave are perpendicular to each other and to the direction of the wave propagation. Electromagnetic waves are used widely in the food processing for destroying microbes.
Therefore, the correct order of electromagnetic waves from the lowest frequency to the highest frequency are gamma rays, X-ray, visible light, and infrared.
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Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below. How much time elapses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately 1530 m/s.)
Answer:
Time, t = 0.104 seconds
Explanation:
Frequency of the click of the Dolphin, f = 55.3 kHz
A dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below, d = 80 m
The speed of sound in seawater is, v = 1530 m/s
Once the sound is send and reflects, the total distance covered by it is 2d such that,
[tex]t=\dfrac{2d}{v}\\\\t=\dfrac{2\times 80}{1530}\\\\t=0.104\ s[/tex]
So, the time elapses before the dolphin hears the echoes of the clicks is 0.104 seconds.
A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.190 m , to a hanging book with mass 3.05 kg . The system is released from rest, and the books are observed to move a distance 1.19 m over a time interval of 0.800 s .
A) What is the tension in the part of the cord attached to the textbook?
B) What is the tension in the part of the cord attached to the book?
C) What is the moment of inertia of the pulley about its rotation axis?
Answer:
a. 7.38 N b. 40.87 N c. 0.113 kg-m²
Explanation:
When the batter hit the foul ball, the baseball moved upward to a delighted fan in the top deck.
A. KE increases and PE increases.
B> KE increases and PE decreases.
C. KE decreases and PE increases.
D. KE decreases and PE decreases.
We can see here that the correct answer is: A. KE increases and PE increases.
As the ball moves upward against gravity, its height increases, indicating a rise in its potential energy (PE). Since the ball moves, its speed increases, implying an increase in its kinetic energy (KE) as well.
KE (kinetic energy) is the energy possessed by an object due to its motion. When the batter hits the foul ball, the baseball is set in motion, and its kinetic energy increases as it moves upward.
PE (potential energy) is the energy possessed by an object due to its position or condition.
Tire inflation is very important to the safe and economical operation of any vehicle. Technician A says that the tire pressure should never exceed the maximum pressure imprinted on the sidewall of the tire. Technician B says to inflate tires to the pressures recommended on the tire information decal or placard on the drivers door. Which technician is correct? Group of answer choices: a) Technician A only b) Tehnician B only c) Both tehnicians A and B d) Neither tehnicians A nor B.
Answer:
The correct option is;
c) Both technicians A and B
Explanation:
Normally the pressure required for a normally inflated tire is included on the side or sidewall of the tire along with the instruction as to the qualification and training required to inflate the vehicle tire. Other information on the sidewall includes the tire model, brand name and manufacturer.
The proper inflation pressure can also be located on the tire information manual or decal as well on the door on the driver's side.
Technician B is correct in recommending that tires should be inflated to the pressure specified on the vehicle's placard. It is more accurate than the maximum pressure imprinted on the tire sidewall.
When considering tire inflation for safe and economical operation of a vehicle, there seem to be conflicting recommendations from two technicians on the correct pressure. Technician A suggests that tire pressure should never exceed the maximum pressure imprinted on the sidewall of the tire. However, this is not necessarily the recommended operating pressure; it is simply the maximum pressure that the tire can safely handle. Technician B provides a more accurate recommendation by advising to inflate tires to the pressures recommended on the tire information decal or placard on the driver's door. This placard usually lists the optimal pressure for the vehicle's tires based on the manufacturer's specifications, and accounts for the best performance and safety.
Most vehicle manufacturers specify the optimal tire pressure for cold tires, which might differ from the maximum pressure on the tire sidewall. Maintaining the proper air pressure in the tires ensures smoother and safer rides, better gas mileage, and longer tire life. Therefore, the correct answer to the question is (b) Technician B only.
Two circular loops of wire, each containing a single turn, have the same radius of 5.00 cm and a common center. The planes of the loops are perpendicular. Each carries a current of 2.10 A. What is the magnitude of the net magnetic field at the common center
The magnitude of the net magnetic field at the common center can be calculated using the formula MOI B = (at the center of the loop)/2R. The magnetic field strength at the center is approximately 6.28 × 10-5 T.
Explanation:The magnitude of the net magnetic field at the common center can be calculated using the formula for the magnetic field strength at the center of a circular loop, which is given by MOI B = (at the center of the loop), 2R. In this case, the radius of each loop is 5.00 cm. Plugging in the values, the magnetic field strength at the center can be calculated.
Substituting the given values into the equation:
Magnetic field strength = MOI B = (2 × π × (5.00 cm)²)(2.10 A)/(2 × 5.00 cm) ≈ 6.28 × 10-5 T
Therefore, the magnitude of the net magnetic field at the common center is approximately 6.28 × 10-5 T.
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The net magnetic field at the common center of the loops will be 3.71 × 10⁻⁵ T.
The two circular loop of wire have their planes perpendicular to each other and share a common centre. This indicates that the direction of net magnetic field at the common centre due to any one of the loops will be perpendicular to the direction of the net magnetic field at the common centre due to the other loop.
As both the loops have the same radius (r), same number of turns (N), and carry the same current (I), hence, the magnetic field at the centre of the loops will have the same magnitude and can be given by the formula:
[tex]B = \frac{N\mu_oI}{2r}[/tex]
Given that:
N = 1
I = 2.10 A
r = 5.00 cm = 0.05 m
μ₀ = 4π × 10⁻⁷
Substituting the values, we get:
[tex]B = \frac{(4 \pi \times 10^{-7})(2.10 A)}{2(0.05 m)}[/tex]
B = 2.63 × 10⁻⁵ T
The net magnetic field at the common centre can be found using
[tex]B_{net} = \sqrt{B_1^{2} +B_2^{2} +2B_1B_2cos\theta}[/tex]
as both the magnetic fields are perpendicular to each other the angle (θ) between them will be 90°.
∴ [tex]B_{net} = \sqrt{B_1^{2} +B_2^{2} }[/tex]
[tex]B_{net} = \sqrt{2B^{2} }[/tex]
[tex]B_{net} = \sqrt{2(2.63 \times 10^{-5} T)^{2} }[/tex]
[tex]B_{net} = (2.63 \times 10^{-5} T)\sqrt{2} }[/tex]
or, [tex]B_{net} = 3.71 \times 10^{-5} \hspace{0.5mm} T[/tex]
PLEASEEEE ANSWERRR FASTT!!! Summarize the contributions that Ellen Swallow Richards made to help
clean up the Earth. *
Answer:
Ellen tested the water in Massachusetts and it had way too much chlorine. She talked to the city and got them to stop dumping waste and garbage into the water.
Explanation:
oh btw im in 5th grade and i get these kind of questions your welcome
Which statement best explains the relationship between current, voltage, and resistance?
A) If we increase the amount of voltage applied, and do not change the resistance, this will result in a decrease in current.
B) If we decrease the current applied, and do not change the resistance, we increase the voltage.
C) If we increase the amount of voltage applied, and do not change the resistance, we will also increase the current.
D) If we decrease the amount of current, this will not affect the amount of voltage, only the amount of resistance.
(FLVS Physical Science 03.09)
Answer:
c
Explanation:
big brain
The heat loss through a window (R-2) is 10 MMBTU/year. Calculate the payback period (in years) ifArgon is filled in the window to increase the effective R-value of the window to 7. Assume heating price to be $10/MMBTU and the cost for filling Argon is $20.Heat lost through new window per year = (10 x 2) / 7 = 2.84 MMBTUs/yrReduction in heat lost per year = 10 - 2.84 = 7.16 MMBTUs/yrSavings in energy cost per year = 7.16 x 10 = $71.6Payback period = 20 / 71.6 = 0.28 yrs
Answer:
0.28 year
Explanation:
Heat lost through new window per year = heat loss through a window / R- value = (10 MMBTU/year × 2) / 7 = 2.86 MMBTUs/yr
Reduction in heat lost per year = heat loss through a window - Heat lost through new window per year = 10 MMBTUs/yr - 2.86 MMBTUs/yr = 7.14 MMBTUs/yr
Savings in energy cost per year = Reduction in heat lost per year × heating price = 7.14 MMBTUs/yr × $10/MMBTU = $71.4 / yr
Payback period = cost for filling Argon / Savings in energy cost per year = $20 / $71.4 /yr = 0.28 yrs
In an inelastic collision, a steel ball of mass 200 g was hit hard into a large ball of dough of mass 700 g. The velocity of the steel ball was 25 m/s. After the collision, the steel ball was stuck in the dough, and this combined system rolled for some time. What was the velocity of the system of the steel ball and dough after the collision?A) 5.6 m/sB) 7.5 m/sC) 8.8 m/sD) 11.1 m/s
Answer:
Velocity after collision will be 5.56 m/sec
So option (a) will be correct answer
Explanation:
Mass of steel ball [tex]m_1=200gram=0.2kg[/tex]
Speed of steel ball before collision [tex]v_1=25m/sec[/tex]
Mass of large ball [tex]m_2=700gram=0.7kg[/tex]
Velocity of large ball [tex]v_2=0m/sec[/tex]
According to conservation of momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]0.2\times 25+0.7\times 0=(0.7+0.2)v[/tex]
[tex]v=5.56m/sec[/tex]
So velocity after collision will be 5.56 m/sec
So option (a) will be correct answer
Which explains earthquakes and volcanic eruptions
Answer: plate tectonics
Explanation: plate tectonics move constantly movement in narrow zones along plate boundaries causes earthquakes and plate tectonics come together and create pressure upon the magma deposit underground forcing it up into a volcanic eruption
A conducting loop is lying flat on the ground. The north pole of a bar magnet is grought down toward the loop. As the magnet approaches the loop, will the magnetic field created by the induced current point up of down?
Answer:
The current in the loop will flow in anticlockwise direction , and the magnetic field created by the induced current point up.
Explanation:
if an electron in an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T west, what are the direction and the magnitude of the velocity?
Answer:
[tex]1.51\cdot 10^6 m/s[/tex] north
Explanation:
When a charged particle moves in a magnetic field, the particle experiences a force given by the formula:
[tex]F=qvB sin \theta[/tex]
where
q is the magnitude of the charge
v is its velocity
B is the magnetic field
[tex]\theta[/tex] is the angle between the directions of v and B
In this problem,
[tex]q=1.6\cdot 10^{-19}C[/tex] (charge of the electron)
[tex]B=8.3\cdot 10^{-2} T[/tex] (strength of magnetic field)
[tex]F=2.0\cdot 10^{-14} N[/tex] (force)
[tex]\theta=90^{\circ}[/tex]
Therefore, the velocity is
[tex]v=\frac{F}{qB sin \theta}=\frac{2.0\cdot 10^{-14}}{(1.6\cdot 10^{-19})(8.3\cdot 10^{-2})(sin 90^{\circ})}=1.51\cdot 10^6 m/s[/tex]
The direction of the force is perpendicular to both the direction of the velocity and the magnetic field, and it can be found using the right-hand rule:
. Thumb: direction of the force (downward) --> however the charge is negative, so this direction must be reversed: upward
- Middle finger: direction of the field (west)
- Index finger: direction of velocity --> north
So, the electron is travelling north.
A stone is thrown vertically upwards with an initial velocity of 30meter per second from the top of a tower 20metes high find the time taken to reach the maximum height and the total time which elapses before it reaches the ground and gravity is 10meter per second .
Answer:
time taken to reach maximum height(t)=3seconds
Explanation:
1st equation of motion:v=u-gt ,but v=0 at maximum height.so 0=u-gt
t=u÷g where,u=30m/s,h=20m,g=10m/s square
so t=30÷10
t=3seconds
If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25.0 m in Lake Michigan (which is fresh water), what will be the volume at the surface of an N2 bubble that occupied 1.00 mm3 in his blood at the lower depth?
Answer:
3.42 cubic mm
Explanation:
Let density of fresh water be [tex]\rho_w = 1000 kg/m^3[/tex]
And atmospheric pressure at the water surface be [tex]P_a = 101325 Pa[/tex]
Let g = 9.8 m/s2. The pressure at the dept of h = 25 m is
[tex]P = P_a + \rho_wgh = 101325 + 1000*9.8*25 = 346325 Pa[/tex]
Using ideal gas law and assume constant temperature, we have the following equation to calculate the volume at the water surface [tex]V_a[/tex], knowing that the volume at the lower depth V = 1 mm3:
[tex]P_aV_a = PV[/tex]
[tex]V_a = PV/P_a = 346325*1/101325 = 3.42 mm^3[/tex]
The volume of a nitrogen bubble in a diver's blood will expand when the pressure decreases as they rise to the surface. If the bubble was 1.00 mm3 at 25.0 m depth, it would expand to 2.5 mm3 at the surface. Rapid ascent can cause the bubbles to expand too quickly, leading to the condition known as Decompression Sickness, or The Bends.
Explanation:The volume expansion of nitrogen bubbles in a diver's blood due to rapid ascent from a certain depth is governed by Boyle's Law that states the volume of gas is inversely proportional to its pressure, assuming the temperature remains constant. When a diver rises quickly from a depth of 25.0 m (approximately 2.5 atm, as each 10m of water is approximately 1 atm) to the surface (1 atm), the pressure decreases. According to Boyle's Law, the volume would therefore increase, thus, if a nitrogen bubble occupied 1.00 mm3 at the depth, at the surface it would expand to 2.5 x 1.00 mm3, or 2.5 mm3. This is why divers must ascend slowly, to allow gases to dissipate gradually and avoid the dangerous condition known as Decompression Sickness or The Bends.
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Does the air exert a buoyant force on all objects in air or only on objects such as balloons that are very light for their size? Does the air exert a buoyant force on all objects in air or only on objects such as balloons that are very light for their size? Air exerts a buoyant force on all objects. Air has too low a density to exert a buoyant force. Air exerts a buoyant force only on objects less dense than air. Air exerts a buoyant force only on objects more dense than air.
Answer:
See explanation
Explanation:
Solution:-
Buoyancy is the force that causes objects to float. It is the force exerted on an object that is partly or wholly immersed in a fluid. Buoyancy is caused by the differences in pressure acting on opposite sides of an object immersed in a static fluid. It is also known as the buoyant force. Buoyancy is the phenomena due to Buoyant Force.
It is as an upward force exerted by a fluid that opposes the weight of an object immersed in a fluid. As we know, the pressure in a fluid column increases with depth. Thus, the pressure at the bottom of an object submerged in the fluid is greater than that at the top. The difference in this pressure results in a net upward force on the object which we define as buoyancy.
- The formula for buoyant force (Fb) is given:
Fb = ρ*g*V
- The force acts on all objects. However, it depends on the fluid density and amount of volume displaced.
- The Buoyant force exerted by air with density = 1.225 kg/m^3 on an object with volume (V) is:
Fb = ρ*g*V = 1.225*9.81*V = 12.02*V
- For the similar object with mass (m), the downward weight would be:
W = m*g
- For the object to float the buoyant force (Fb) must be greater than weight of the object:
Fb > W
12.02*V > m*9.81
V / m > 0.816
- The ratio of V / m must be at-least = 0.816.
- Assuming the object is fully immersed in air, then the volume displaced V = ρ_material*V
ρ_material < 1 / 0.816
ρ_material < 1.225 or ( ρ_air )
- So the for an object to float in air its material density must always be less than that of air. That why in balloons lighter gas is used which have density less than that of air like Helium.
Final answer:
The air exerts a buoyant force on all objects due to Archimedes' principle, which states that the force on an object in a fluid is equal to the weight of the fluid it displaces. An object's ability to float is determined by its density relative to the surrounding air. All objects are subject to this force, whether they rise, sink, or remain suspended.
Explanation:
The air does indeed exert a buoyant force on all objects, not just on light objects like balloons. This buoyant force is a result of Archimedes' principle, which states that the force on an object in a fluid is equal to the weight of the fluid it displaces. Therefore, if an object's average density is less than that of the surrounding fluid, it will float. On the other hand, if an object's density is greater, it will sink. The key factor is the object's density relative to the air around it, determining if it will rise, fall, or remain suspended.
This principle extends to all objects in fluids, including gases such as air. It means that even objects that do not float, such as rocks or metal items, experience a buoyant force; however, this force is not enough to overcome their weight and make them float. Similarly, our bodies are buoyed by the atmosphere to some degree; it's just that the effect is not as noticeable as it is with helium balloons, which have a density much lower than that of air.
An apple in a tree has a gravitational store of 8J. As it falls, it accelerates constantly until it hits the ground. What is the apple’s maximum kinetic energy store? Explain your answer.
Answer:
Explanation:
Given that on the tree the gravitational energy stored is 8J
Then, mgh = 8J.
The apple begins to fall and hit the ground, what is the maximum kinetic energy?
Using conservation of energy, as the above is about to hit the ground, the apple is at is maximum speed, and the height then is 0m, so the potential energy at the ground is zero, so all the potential of the apple at the too of the tree is converted to kinetic energy as it is about to hits the ground. Along the way to the ground, both the Kinetic energy and potential energy is conserved, it is notice that at the top of the tree, the apple has only potential energy since velocity is zero at top, and at the bottom of the tree the apple has only kinetic energy since potential energy is zero(height=0)
So,
K.E(max) = 8J
At 20 ◦C, a rod is exactly 20.05 cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 270 ◦C, where the rod now measures 20.11 cm on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made? Take αsteel = 1.2 × 10−5 K−1 .
The coefficient of linear expansion for the material of the rod can be calculated using the formula ΔL/L = α ΔT to be α = 2 x 10⁻⁵ °C⁻¹.
Explanation:The subject of this question is the calculation of the coefficient of linear expansion for a certain material. The relevant formula here is: ΔL/L = α ΔT, where ΔL is change in length, L is original length, α is coefficient of linear expansion and ΔT is change in temperature.
Firstly, we calculate the change in length of the rod (ΔL) which is the final length minus initial length, 20.11 cm - 20.05 cm = 0.06 cm. Then, the change in temperature (ΔT) is the final temperature minus the initial one: 270°C - 20°C = 250°C.
After substitution into the formula, we get 0.06 cm / 20.05 cm = α × 250°C. Solving for α, we get the coefficient of linear expansion for the material of the rod to be α = 2 x 10-5 °C⁻¹.
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The coefficient of linear expansion for the material of the rod is approximately 0.011974 cm/°C.
Explanation:Solution:To find the coefficient of linear expansion for the material of which the rod is made, we can use the equation for linear thermal expansion:
ΔL = αLΔT
where ΔL is the change in length, α is the coefficient of linear expansion, L is the initial length of the rod, and ΔT is the change in temperature.
Given that the initial length of the rod is 20.05 cm, the final length is 20.11 cm, and the change in temperature is 270 °C - 20 °C = 250 °C, we can substitute these values into the equation to solve for α:
ΔL = αLΔT20.11 cm - 20.05 cm = (α)(20.05 cm)(250 °C)0.06 cm = 5.0125 α °Cα = 0.06 cm / 5.0125 °Cα = 0.011974 cm/°CHence, the coefficient of linear expansion for the material of which the rod is made is approximately 0.011974 cm/°C.
Learn more about linear expansion here:https://brainly.com/question/35554239
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A Swinging Monkey A 8.72 kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume that the radial distance between the branch and the point where the monkey's mass is located to be 84.1 cm. As it swings through the lowest point on the circle, the monkey has a speed of 2.53 m/s. Calculate the magnitude of the centripetal force acting on the monkey.
Answer:
Centripetal force, F = 66.36 N
Explanation:
Given that,
Mass of the monkey, m = 8.72 kg
The radial distance between the branch and the point where the monkey's mass is located to be 84.1 cm, r = 84.1 cm
The speed of the monkey, v = 2.53 m/s
We need to find the magnitude of the centripetal force acting on the monkey. It is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{8.72\times (2.53)^2}{84.1\times 10^{-2}}\\\\F=66.36\ N[/tex]
So, the magnitude of the centripetal force acting on the monkey is 66.36 N.
An ideal gas with k 5 1.4 is flowing through a nozzle such that the Mach number is 1.8 where the flow area is 36 cm2. Approximating the flow as isentropic, determine the flow area at the location where the Mach number is 0.9.
Answer:
The flow area at the location where the Mach number is 0.9 is 25.24 cm²
Explanation:
Here we have for isentropic flow;
[tex]\frac{A}{A^*} = \frac{1}{M}(\frac{2}{k+1} (1+\frac{k-1}{2}M^2))^{(\frac{k+1}{2(k-1)} )[/tex]
Where:
A = Area of flow = 36 cm²
M = Mach number at section of = 1.8
k = Specific heat ratio = 1.4
A* = Area at the throat
Therefore, plugging the values we get
[tex]\frac{36}{A^*} = \frac{1}{1.8}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}1.8^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.439[/tex]
Therefore, A* = 36/1.439 = 25.01769 cm²
Where the Mach number is 0.9, we have
[tex]\frac{A}{25.02} = \frac{1}{0.9}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}0.9^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.009[/tex]
Therefore A = 25.020× 1.009 = 25.24 cm²
The flow area at the location where the Mach number is 0.9 = 25.24 cm².
Where is the image located in the case of an object placed at a distance to the left of a converging lens?
Answer:
the object is placed at 2f
Explanation:
Converging lens are convex lens, as the name emanates , they are light rays that converge at a point via a certain distance at the opposite side. Often times , any incident rays that is moving and is parallel to the principal axis is usually termed as Converging lens.
So when the object is placed at 2f , the image will be formed at 2f to the right for converging lens.
In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 4.7 rev/s in 8.3 s. You lift the lid of the washer and notice that the tub decelerates and comes to a stop in 17.5 s. Assuming that the tub rotates with constant angular acceleration while it is starting and stopping, determine the total number of revolutions undergone by the tub during this entire time interval.
Answer:
Number of revolutions = Δθ = 60.63 revs
Explanation:
The angular acceleration of the washer is given by
α = (ωf - ωi)/Δt
Where ωf is the initial angular speed, ωi is the final angular speed and Δt is the interval of time during this acceleration.
α = (4.7 - 0)/8.3
α = 0.56 rev/s²
As we know from the equation of kinematics,
2αΔθ = ωf² - ωi²
Where Δθ is the change in angular displacement of the washer.
Δθ = (ωf² - ωi²)/2α
Δθ = (4.7² - 0²)/2*0.56
Δθ = 19.72 revs
Now the tub decelerates and comes to a stop in 17.5 s
α = (ωf - ωi)/Δt
α = (-4.7 - 0)/17.5
α = -0.27 rev/s²
the corresponding change in angular displacement of the washer is
Δθ = (ωf² - ωi²)/2α
Δθ = (0² - 4.7²)/2*-0.27
Δθ = 40.91 revs
Therefore, the total number of revolutions undergone by the tub during this entire time interval is
Δθ = 19.72 + 40.91
Δθ = 60.63 revs
What is the force exerted on a moving charge of -2.0 C at a 20° angle through a magnetic field of 3.0 x 10-4T with a
velocity of 5.0 x 106 m/s?
Answer:
1026 N
Explanation:
Force on a moving charge in a magnetic field is given by
F= qvB Sinθ
F = 2 x 5 x 10^6 x 3 x 10-4 x Sin (20)
F = 1026 N
Answer:
1.0 x 10^-3 N
Explanation:
got it right on edge:3
A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Do you think it would be reasonable to use this polynomial to estimate the distance traveled during the next minute? Why or why not?
Answer:
[tex]T(1)=21 [/tex]
Explanation:
The equation of the position in kinematics is given:
[tex]x(t)=x_{0}+v_{0}t+0.5at^{2}[/tex]
x(0) is the initial position, in this it is 0 v(0) is the initial velocity (20 m/s)a is the acceleration (2 m/s²)So the equation will be:
[tex]x(t)=20t+0.5*2*t^{2}[/tex]
[tex]x(t)=20t+t^{2}[/tex]
Now, the Taylor polynomial equation is:
[tex]f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...[/tex]
Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.
[tex]f(t)=x(t)=20t+t^{2}[/tex]
[tex]f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t[/tex]
[tex]f''(t)=\frac{dv(t)}{dt}=a(t)=2[/tex]
Using the Taylor polynomial with a = 0 and take just the second order of the derivative.
[tex]f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}[/tex]
[tex]f(0)=x(0)=0[/tex]
[tex]f'(0)=v(0)=20[/tex]
[tex]f''(0)=a(0)=2[/tex]
[tex]T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}[/tex]
[tex]T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}[/tex]
[tex]T(t)=20t+t^{2}[/tex]
Let's put t=1 so find the how far the car moves in the next second:
[tex]T(1)=20*1+1^{2}[/tex]
[tex]T(1)=21 [/tex]
Therefore, the position in the next second is 21 m.
We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.
I hope it helps you!