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2 years ago

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A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this

time
find distance and acceleration

PLEASEEEE HELP I HAVE A PHYSICS EXAM TOMORROW AND IM STRUGGLING

1 answer:

krok68 [10]2 years ago

8 0

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

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A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

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Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

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At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

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$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

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Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

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