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Greeley [361]
3 years ago
11

A stone weighing 1.5 kilograms is resting on a rock at a height of 20 meters above the ground. The stone rolls down 10 meters an

d comes to rest on a patch of moss. The gravitational potential energy of the stone on the moss is joules. (Use PE = m × g × h, where g = 9.8 N/kg.)
Physics
2 answers:
Andrews [41]3 years ago
6 0
The potential energy of the moss at its position is 147 Joules. 
34kurt3 years ago
4 0
Given that the equation for the potential energy of an object is: PE = mgh, where, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground, simply substitute the given values to obtain the potential energy. In this case, m = 1.5 kg, g = 9.8 N/kg, and h = 10 m. The height is set to 10 m since this is the height of the moss above the ground after it fell. Thus, the potential energy of the moss at its position is 147 Joules. 
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Answer:

131.25

Explanation:

i worked it out on a diffrent sheet so its hard to explain

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Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0
4 years ago
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egoroff_w [7]

Answer:

the smallest radius of the circular path is 8.1 km

Explanation:

The computation of the smallest radius of the circular path is given below:

Given that

V = Velocity = 201 m/s

a_c = acceleration = 5 m/s^2

radius = ?

As we know that

a_c = V^2 ÷ r

5 = 201^2 ÷ r

r = 201^2 ÷ 5

= 8,080.2 g

= 8.1 km

Hence, the smallest radius of the circular path is 8.1 km

4 0
3 years ago
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yanalaym [24]
The answer is true I think
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