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Sphinxa [80]
3 years ago
7

A ball is thrown vertically downward with an initial speed of 12.0 m/s from a height of 10.0 m above the ground. Ignoring air re

sistance, what is the speed of the ball when it reaches the ground?
(A) 18.5 m/s
(B) 14.6 m/s
(C) 14.0 m/s
(D) 12.8 m/s
(E) 12.0 m/s
Can you just take a picture of your work clearly because I have to show work. Thanks.
Physics
1 answer:
natita [175]3 years ago
8 0

Answer:

18.5 m/s

Explanation:

Take down to be positive.

Given:

Δy = 10.0 m

v₀ = 12.0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (12.0 m/s)² + 2 (10 m/s²) (10.0 m)

v = 18.5 m/s

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the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

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Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

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8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

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6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

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