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pychu [463]
3 years ago
10

A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s

Physics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

v^2-u^2=2as

Applying values as:

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s

<u>The speed with which the ball hit the ground = 38.87 m/s</u>

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Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

<u>ΔL = 0.66 m</u>

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