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Tomtit [17]
3 years ago
15

A brick is dropped from rest from the top of a building through air (air resistance is present) to the ground below. how does th

e brick's kinetic energy just before reaching the ground compare with the gravitational potential energy at the top of the building? set zero height at the ground level. the comparison cannot be made since the information of air resistance is not given. kinetic energy just before reaching the ground < gravitational potential energy at the top of the building kinetic energy just before reaching the ground = gravitational potential energy at the top of the building kinetic energy just before reaching the ground > gravitational potential energy at the top of the building
Physics
1 answer:
sasho [114]3 years ago
5 0
The correct option is this: KINETIC ENERGY JUST BEFORE REACHING THE GROUND IS LESS THAN THE GRAVITATIONAL POTENTIAL ENERGY AT THE TOP OF THE BUILDING.
When the object was on the top of the building, it has potential energy. This potential energy was converted to kinetic energy when the object started falling. In the process of falling, friction was present, which means that some of the energy will be converted to heat as a result of the friction. Therefore the kinetic energy of the falling object will be less than its potential energy, because some of the energy has been spent on friction.<span />
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A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

41.7=1.12 u +4.9 \times 1.12^{2}

u = 31.75 m/s

5 0
3 years ago
What occurs along a convergent plate boundary?
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3 years ago
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2. None of these occur

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The layer of earth that has the lightest elements is the
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If I remember correctly (from my studies long time ago) the layers are from the outer to the center:
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8 0
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