I believe that for 1 the answer is 5N. Work = Force * Distance and since we have two of those variables we can figure out the other one. We can substitute 40 J for W and 8 m for D.
40J = F * 8m
Then we isolate F so we can solve for it.
40J/8m = F
5N = F
1. In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. The average force exerted is?
my answer: 320N
2. What is the kinetic energy of a 20 kg object moving at a speed of 10 m/s?
my answer: 1000J
3. the gravitational potential energy of a bird of mass 2.0 kg on a tree branch of height 15 m is most nearly
my answer: 300J
4. Two objects have the same mass. One is travelling twice as fast as the other. The work that must be done to stop the faster object compared to the work required to stop the slower object is:
a)two times greater
b)the same
c)four times greater
d)half as great
e)one quarter as great
my answer: a
5. When a rock is thrown straight up in the air, after it leaves the hand, the rock begins to slow down. This occurs because:
a) the rock is gaining potential energy as it rises and thus it must lose kinetic energy
b)the force of gravity acting on the rock increases as the rock rises
c)the forces acting on the rock are balanced
d)the potential energy of the rock decreases as the rock rises
my answer: d
45 answers
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. The average force exerted is
5N.
What is the kinetic energy of a 20 kg object moving at a speed of 10 m/s?
The kinetic energy (KE) of an object with mass (m) moving at a velocity (v) is given by the equation:
KE = 1/2 * m * v^2
Plugging in the given values, we get:
KE = 1/2 * 20 kg * (10 m/s)^2
KE = 1000 J
Therefore, the kinetic energy of the 20 kg object moving at 10 m/s is 1000 J.
KE = 1/2 * m * v^2
Plugging in the given values, we get:
KE = 1/2 * 20 kg * (10 m/s)^2
KE = 1000 J
Therefore, the kinetic energy of the 20 kg object moving at 10 m/s is 1000 J.
The gravitational potential energy of a bird of mass 2.0 kg on a tree branch of height 15 m is closest to
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. If the kinetic energy gained by the mass is 34 J what is the WORK DONE by the NON CONSERVATIVE force?
We can solve for the work done by the non-conservative force using the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. Thus, we have:
Net work = Change in kinetic energy
Total work - Non-conservative work = Kinetic energy gained
40 J - Non-conservative work = 34 J
Non-conservative work = 6 J
Therefore, the work done by the non-conservative force is 6 J.
Net work = Change in kinetic energy
Total work - Non-conservative work = Kinetic energy gained
40 J - Non-conservative work = 34 J
Non-conservative work = 6 J
Therefore, the work done by the non-conservative force is 6 J.
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. If the kinetic energy gained by the mass is 34 J what is the EFFICIENCY?
The efficiency (η) is defined as the ratio of the output energy to the input energy, expressed as a percentage. In this case, the output energy is the kinetic energy gained by the mass (34 J), and the input energy is the work done on the mass (40 J). Thus:
η = (output energy / input energy) x 100%
η = (34 J / 40 J) x 100%
η = 0.85 x 100%
η = 85%
Therefore, the efficiency is 85%.
η = (output energy / input energy) x 100%
η = (34 J / 40 J) x 100%
η = 0.85 x 100%
η = 85%
Therefore, the efficiency is 85%.
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done in 4 s. What is the POWER of the work done by the force.
The power (P) is defined as the rate at which work is done, and is given by the equation:
P = W / t
where W is the work done and t is the time taken. So in this case, we have:
W = 40 J
t = 4 s
Substituting these values, we get:
P = 40 J / 4 s
P = 10 watts
Therefore, the power of the work done by the force is 10 watts.
P = W / t
where W is the work done and t is the time taken. So in this case, we have:
W = 40 J
t = 4 s
Substituting these values, we get:
P = 40 J / 4 s
P = 10 watts
Therefore, the power of the work done by the force is 10 watts.
The gravitational potential energy of a bird of mass 2.0 kg on a tree branch of height 15 m is closest to...?
The gravitational potential energy (PE) of an object with mass (m) at a height (h) above a reference level is given by the equation:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2 at the Earth's surface).
Plugging in the given values, we get:
PE = 2.0 kg * 9.8 m/s^2 * 15 m
PE = 294 J
Therefore, the gravitational potential energy of the bird is closest to 294 J.
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2 at the Earth's surface).
Plugging in the given values, we get:
PE = 2.0 kg * 9.8 m/s^2 * 15 m
PE = 294 J
Therefore, the gravitational potential energy of the bird is closest to 294 J.
A 100 kg mass is dropped from a height of 3.0 m. The kinetic energy before striking the ground is 2500J. What is the efficiency of the fall?
We can use the conservation of energy principle to find the efficiency of the fall. The initial potential energy of the mass is converted into kinetic energy as it falls, neglecting air resistance. Therefore:
Initial potential energy = mgh
= (100 kg)(9.8 m/s^2)(3.0 m)
= 2940 J
Efficiency = (kinetic energy before striking the ground / initial potential energy) x 100%
= (2500 J / 2940 J) x 100%
= 85%
Therefore, the efficiency of the fall is 85%.
Initial potential energy = mgh
= (100 kg)(9.8 m/s^2)(3.0 m)
= 2940 J
Efficiency = (kinetic energy before striking the ground / initial potential energy) x 100%
= (2500 J / 2940 J) x 100%
= 85%
Therefore, the efficiency of the fall is 85%.
A 100 kg mass is dropped from a height of 3.0 m. If the efficiency is 60% what is the kinetic energy of the mass upon impact?
If the efficiency of the fall is 60%, it means that only 60% of the initial potential energy is converted into kinetic energy upon impact. The rest is lost to non-conservative forces like air resistance, sound, and heat.
The initial potential energy of the mass is:
PE = mgh
= (100 kg)(9.8 m/s^2)(3.0 m)
= 2940 J
Due to the 60% efficiency, the kinetic energy of the mass upon impact is:
KE = efficiency x PE
= 0.60 x 2940 J
= 1764 J
Therefore, the kinetic energy of the mass upon impact is 1764 J.
The initial potential energy of the mass is:
PE = mgh
= (100 kg)(9.8 m/s^2)(3.0 m)
= 2940 J
Due to the 60% efficiency, the kinetic energy of the mass upon impact is:
KE = efficiency x PE
= 0.60 x 2940 J
= 1764 J
Therefore, the kinetic energy of the mass upon impact is 1764 J.
A 100 kg mass is dropped from a height of 3.0 m. If the efficiency is 60% what is the velocity of the mass upon impact?
The efficiency of the system is defined as the ratio of kinetic energy after the fall to the potential energy before the fall. So, we can solve for the kinetic energy of the mass when it hits the ground using the efficiency formula:
Efficiency = Kinetic energy after the fall / Potential energy before the fall
We know the mass is 100 kg, the height is 3.0 m, and g is 9.8 m/s^2. Therefore, the potential energy before the fall is:
Potential Energy before the fall = mgh
= (100 kg)(9.8 m/s^2)(3.0 m)
= 2940 J
Now we can solve for the kinetic energy after the fall:
Efficiency = Kinetic energy after the fall / Potential energy before the fall
0.60 = Kinetic energy after the fall / 2940 J
Kinetic energy after the fall = 0.60 * 2940 J
Kinetic energy after the fall = 1764 J
We can calculate the velocity of the mass just before it hits the ground, using the kinetic energy formula:
Kinetic energy = 1/2 * m * v^2
Solving for v, we get:
v = sqrt(2 * kinetic energy / m)
v = sqrt(2 * 1764 J / 100 kg)
v = 26.5 m/s
Therefore, the velocity of the mass just before it hits the ground is about 26.5 m/s (rounded to one decimal place).
Efficiency = Kinetic energy after the fall / Potential energy before the fall
We know the mass is 100 kg, the height is 3.0 m, and g is 9.8 m/s^2. Therefore, the potential energy before the fall is:
Potential Energy before the fall = mgh
= (100 kg)(9.8 m/s^2)(3.0 m)
= 2940 J
Now we can solve for the kinetic energy after the fall:
Efficiency = Kinetic energy after the fall / Potential energy before the fall
0.60 = Kinetic energy after the fall / 2940 J
Kinetic energy after the fall = 0.60 * 2940 J
Kinetic energy after the fall = 1764 J
We can calculate the velocity of the mass just before it hits the ground, using the kinetic energy formula:
Kinetic energy = 1/2 * m * v^2
Solving for v, we get:
v = sqrt(2 * kinetic energy / m)
v = sqrt(2 * 1764 J / 100 kg)
v = 26.5 m/s
Therefore, the velocity of the mass just before it hits the ground is about 26.5 m/s (rounded to one decimal place).
A 100 kg mass is dropped from a height of 3.0 m. If the efficiency is 60% what is the velocity of the mass upon impact?
a 1 point
b 5.9 m/s
c 11 m/s
d 4.85 m/s
e 7.7 m/s
a 1 point
b 5.9 m/s
c 11 m/s
d 4.85 m/s
e 7.7 m/s
The correct answer is (c) 11 m/s.
We can calculate the velocity of the mass just before it hits the ground, using the kinetic energy formula:
Kinetic energy = 1/2 x m x v^2
We know the mass is 100 kg, and the efficiency is 60% which means kinetic energy after the fall is equal to 60% of the potential energy before the fall. So, we can get the potential energy before the fall using the formula:
Potential energy = m x g x h
where g is the acceleration due to gravity (9.8 m/s^2) and h is the initial height (3.0 m).
Potential energy = 100 x 9.8 x 3.0 = 2940 J
Kinetic energy = (60/100) x 2940 J = 1764 J
Now, we can plug in the values in the kinetic energy formula and solve for velocity:
1764 J = 1/2 x 100 kg x v^2
v^2 = 35.28
v = sqrt(35.28) = 5.94 m/s (approx.)
Therefore, the velocity of the mass just before it hits the ground is about 5.94 m/s (rounded to two decimal places). The closest option is (c) 11 m/s.
We can calculate the velocity of the mass just before it hits the ground, using the kinetic energy formula:
Kinetic energy = 1/2 x m x v^2
We know the mass is 100 kg, and the efficiency is 60% which means kinetic energy after the fall is equal to 60% of the potential energy before the fall. So, we can get the potential energy before the fall using the formula:
Potential energy = m x g x h
where g is the acceleration due to gravity (9.8 m/s^2) and h is the initial height (3.0 m).
Potential energy = 100 x 9.8 x 3.0 = 2940 J
Kinetic energy = (60/100) x 2940 J = 1764 J
Now, we can plug in the values in the kinetic energy formula and solve for velocity:
1764 J = 1/2 x 100 kg x v^2
v^2 = 35.28
v = sqrt(35.28) = 5.94 m/s (approx.)
Therefore, the velocity of the mass just before it hits the ground is about 5.94 m/s (rounded to two decimal places). The closest option is (c) 11 m/s.
A girl wants to slide down a FRICTIONLESS playground slide. If she starts at rest from 'X' and slides down to 'Y', which of the following slide shapes will have the highest efficiency? All the slides have the same vertical height and THEY ARE FRICTIONLESS (neglect any air resistance) !!!
Captionless Image
They all have the same efficiency because energy is not lost to friction
A because the time to reach the bottom is the longest.
B because this will be the fastest to the bottom!!!
D because there is more ramp (longer path length) to remove energy as it slides under the force of friction
No way to tell
Captionless Image
They all have the same efficiency because energy is not lost to friction
A because the time to reach the bottom is the longest.
B because this will be the fastest to the bottom!!!
D because there is more ramp (longer path length) to remove energy as it slides under the force of friction
No way to tell
All the slides have the same efficiency because energy is not lost to friction. The efficiency of the slide does not depend on the shape of the slide, since all the slides have the same vertical height and are frictionless. When the girl slides down, the potential energy she gains at the top is converted entirely into kinetic energy at the bottom, neglecting any air resistance. Therefore, the shape of the slide will not affect the efficiency of the slide.
In low earth orbit the atmosphere is extremely thin (very spaced out particles) but reaches 500 Celsius. Therefore this is example of
In low Earth orbit, the atmosphere is extremely thin, but it reaches high temperatures due to the intense solar radiation and the high-speed collisions between particles. This is an example of thermal radiation. Thermal radiation is the emission of electromagnetic waves from a material due to its temperature. The temperature of the particles in the thin atmosphere in low Earth orbit is very high, and they emit thermal radiation in the form of infrared radiation. This radiation can be absorbed by objects in space, causing their temperatures to rise as well.
A thermos bottle is coated with a shiny silver surface on the inside is an example of reducing loss of heat transfer from _______due to internal reflection.
A thermos bottle is coated with a shiny silver surface on the inside, which is an example of reducing loss of heat transfer from radiation due to internal reflection. The shiny silver surface reflects thermal radiation (infrared radiation) back into the thermos bottle, reducing the rate of heat transfer by radiation. This helps to keep hot liquids hot and cold liquids cold for longer periods of time. The shiny silver surface acts as a reflective barrier that prevents heat from being transferred by radiation in both directions between the hot or cold inner contents and the outside environment.
Water has specific heat of 4200 J per kg per degree Celsius. This means that 2.0 kg of water will release 8400 J of energy when it cools down by 1 C.
a) true
b) false
a) true
b) false
The given statement is false. Rather, 2.0 kg of water will require 8400 J of energy to be heated by 1 degree Celsius, according to the specific heat capacity of water given. The specific heat capacity of water tells us how much energy is required to heat up or cool down water by 1 degree Celsius per unit mass. It is not the amount of energy released by water when it cools down by 1 degree Celsius.
Water has specific heat of 4200 J per kg per degree Celsius. This means that 1.0 kg of water will release 8400 J of energy when it cools down by 2 C.
a) true
b) false
C) can not tell
a) true
b) false
C) can not tell
The given statement is false.
The specific heat capacity of water tells us how much energy is required to heat up or cool down water by 1 degree Celsius per unit mass. Given that the specific heat capacity of water is 4200 J/kg°C, we can find the amount of energy required to cool down 1 kg of water by 2 degrees Celsius as follows:
q = m * c * ΔT
where q is the amount of energy required, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Substituting the given values, we get:
q = (1 kg) * (4200 J/kg°C) * (2°C)
q = 8400 J
Therefore, 1.0 kg of water will require 8400 J of energy to be cooled down by 2°C, but it will not release this amount of energy. So the statement is false.
The specific heat capacity of water tells us how much energy is required to heat up or cool down water by 1 degree Celsius per unit mass. Given that the specific heat capacity of water is 4200 J/kg°C, we can find the amount of energy required to cool down 1 kg of water by 2 degrees Celsius as follows:
q = m * c * ΔT
where q is the amount of energy required, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Substituting the given values, we get:
q = (1 kg) * (4200 J/kg°C) * (2°C)
q = 8400 J
Therefore, 1.0 kg of water will require 8400 J of energy to be cooled down by 2°C, but it will not release this amount of energy. So the statement is false.
What is the specific heat capacity of a 2.0 kg metal needing 14000 J to change temperature from 5 C to 6.5 C
We can use the specific heat capacity equation to find the specific heat capacity (c) of the metal, given that it has a mass of 2.0 kg and requires 14000 J of energy to change temperature from 5°C to 6.5°C:
Q = mcΔT
where Q is the amount of energy transferred as heat, m is the mass of the metal, ΔT is the change in temperature, and c is the specific heat capacity.
Substituting the given values, we get:
14000 J = (2.0 kg) * c * (6.5°C - 5°C)
Simplifying the right-hand side, we get:
14000 J = (2.0 kg) * c * (1.5°C)
Dividing both sides by (2.0 kg) * (1.5°C), we get:
c = 14000 J / [(2.0 kg) * (1.5°C)]
c = 4666.67 J/(kg°C)
Therefore, the specific heat capacity of the metal is 4666.67 J/(kg°C).
Q = mcΔT
where Q is the amount of energy transferred as heat, m is the mass of the metal, ΔT is the change in temperature, and c is the specific heat capacity.
Substituting the given values, we get:
14000 J = (2.0 kg) * c * (6.5°C - 5°C)
Simplifying the right-hand side, we get:
14000 J = (2.0 kg) * c * (1.5°C)
Dividing both sides by (2.0 kg) * (1.5°C), we get:
c = 14000 J / [(2.0 kg) * (1.5°C)]
c = 4666.67 J/(kg°C)
Therefore, the specific heat capacity of the metal is 4666.67 J/(kg°C).
Why does a metal FEEL colder to the touch than rubber in the same room?
a) rubber has a much lower heat capacity so feels warmer
b) hey are do not feel different...but the metal is colder
c) the temperature is different but the thermal energy is the same
d) heat capacities are very different ... the metal moves the heat faster from your hand
a) rubber has a much lower heat capacity so feels warmer
b) hey are do not feel different...but the metal is colder
c) the temperature is different but the thermal energy is the same
d) heat capacities are very different ... the metal moves the heat faster from your hand
The correct answer is (d) heat capacities are very different, so the metal moves the heat faster from your hand.
Metal feels colder to the touch than rubber in the same room because metal has a much lower specific heat capacity than rubber. Specific heat capacity is the amount of energy required to raise the temperature of a material by 1 degree Celsius per unit mass. Because of this, metal can transfer heat more quickly from your hand than rubber can, making it feel colder. Metal is an excellent conductor of heat, while rubber is a poor conductor of heat. When you touch metal, it quickly conducts heat away from your hand, making it feel colder, while rubber does not conduct heat away as quickly, so it feels warmer to the touch.
Metal feels colder to the touch than rubber in the same room because metal has a much lower specific heat capacity than rubber. Specific heat capacity is the amount of energy required to raise the temperature of a material by 1 degree Celsius per unit mass. Because of this, metal can transfer heat more quickly from your hand than rubber can, making it feel colder. Metal is an excellent conductor of heat, while rubber is a poor conductor of heat. When you touch metal, it quickly conducts heat away from your hand, making it feel colder, while rubber does not conduct heat away as quickly, so it feels warmer to the touch.
What is the specific heat capacity of a 2.0 kg metal needing 14000 J to change temperature from 5 C to 6.5 C?
a) 9300 J/kgC
b) 4700 J/kgC
c) 1400 J/kgC
d) 7000 J/kgC
a) 9300 J/kgC
b) 4700 J/kgC
c) 1400 J/kgC
d) 7000 J/kgC
The specific heat capacity of the metal is:
c = Q / (m * ΔT)
where Q is the amount of heat transferred, m is the mass of the metal, and ΔT is the change in temperature.
Substituting the given values, we have:
c = 14000 J / (2.0 kg * (6.5°C - 5.0°C))
c = 14000 J / (2.0 kg * 1.5°C)
c = 4666.67 J/(kg°C)
Therefore, the specific heat capacity of the metal is 4666.67 J/(kg°C).
The closest option is (b) 4700 J/kgC.
c = Q / (m * ΔT)
where Q is the amount of heat transferred, m is the mass of the metal, and ΔT is the change in temperature.
Substituting the given values, we have:
c = 14000 J / (2.0 kg * (6.5°C - 5.0°C))
c = 14000 J / (2.0 kg * 1.5°C)
c = 4666.67 J/(kg°C)
Therefore, the specific heat capacity of the metal is 4666.67 J/(kg°C).
The closest option is (b) 4700 J/kgC.
Water has specific heat of 4200 J per kg per degree Celsius. This means that 1.0 kg of water will release 8400 J of energy when it cools down by 2 C.
A) True
B) can't tell I need to know the temperature in kelvin
C) False
A) True
B) can't tell I need to know the temperature in kelvin
C) False
The statement is false.
The specific heat capacity of water, 4200 J/(kg °C), tells us the amount of heat energy required to raise the
The specific heat capacity of water, 4200 J/(kg °C), tells us the amount of heat energy required to raise the
Water has specific heat of 4200 J per kg per degree Celsius. This means that 2.0 kg of water will release 8400 J of energy when it cools down by 1 C.
A) True
B) can't tell I need to know the temperature in kelvin
C) False
A) True
B) can't tell I need to know the temperature in kelvin
C) False
Which has more thermal energy, 1kg of copper at 100 Celsius or 100 Kelvin?
a) 100 K
b) 100 C
c) depends on the material
d) They have the same
a) 100 K
b) 100 C
c) depends on the material
d) They have the same
If you kept the mass and starting temperature of each metal (Silver 234, Gold 128, Copper 385, Iron 452) and the mass and starting temperature of the water constant (water 4200), which sample of hot metal would generate the greatest temperature change in the water? Verify your guess.
How much ice (-10C) would I need to add to 500 g of water to cool the water to 0C from 60c?
hree steel spheres have masses of 12.62g,13.0g and 12.88g. find the average masss of the steel spheres
1 answer