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Vesna [10]
2 years ago
5

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

tion on the crate compared to your push
Physics
1 answer:
poizon [28]2 years ago
3 0

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

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Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha
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Answer:

(i) W = 8.918 N

(ii) V = 9.1 \times 10^{-4} m^3

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

W = mg

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here we know that

\rho = 0.91 g/cm^3

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now the mass of the ice cube is given as

m = 0.91 \times 1000 = 910 g

now weight is given as

W = 0.910 \times 9.8 = 8.918 N

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

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W_{displaced} = 8.918 N

So here volume displaced is given as

\rho_{water}Vg = 8.918

1000(V)9.8 = 8.918

V = 9.1 \times 10^{-4} m^3

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

W = \rho_{water} (L^2 d) g

8.918 = 1000(0.10^2 \times d) 9.8

d = 0.091 m

so it is submerged by d = 9.1 cm inside water

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