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3 years ago

A bowling ball of 35.2kg, generates 218 kg.M/s units of momentum. What is the velocity of the bowling ball?

Chemistry

1 answer:

erik [133]3 years ago

7 0

Answer is: velocity of the bowling ball is 6.19 m/s.

Momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction in three-dimensional space.

m(ball) = 35.2 kg; mass of a bowling ball.

p(ball) = 218 kg·m/s; momentum of a ball.

v(ball) = p(ball) ÷ m(ball).

v(ball) = 218 kg·m/s ÷ 35.2 kg.

v(ball) = 6.19 m/s; velocity of a bowling ball.

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A 10-liter container has 2 moles of oxygen at a pressure of 92 kpa. The effective speed (rms) of the oxygen molecules in the gas

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The effective speed (rms) of the oxygen gas is 293.68 m/s.

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Root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules

$v_{rms}=\sqrt[]{\frac{3RT}{M} }$

<em>where </em>R = universal gas constant

M = molar mass of the gas in kg/mol

T = temperature in Kelvin

According to the ideal gas law,

PV = nRT

RT = $\frac{PV}{n}$

Substitute in the rms velocity formula,

$v_{rms} = \sqrt[]{\frac{3PV}{nM} }$

P = 92 kPa, V = 10 L, n = 2 moles and M = 32 x 10⁻³ kg/mol

$v_{rms} = \sqrt[]{\frac{3\times92\times10}{2\times32\times10^-^3} }$

=293.68 m/s

Thus, the effective speed (rms) of O₂ gas is 293.68 m/s.

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