Question #ee704

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Question #ee704

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Answer 1 · 2017-05-18T23:22:28

4m/s

Explanation:

you have some $\frac{k g \cdot m}{s}$ $(kg*m)/s$ . To get $\frac{m}{s}$ $m/s$ you have to divide by kg:
$\frac{\frac{k g \cdot m}{s}}{k g} = \frac{m \cdot k g}{s \cdot k g} = \frac{m}{s}$ $((kg*m)/s)/(kg)=(m*kg)/(s*kg)=m/s$
For your situation:
$\frac{180 \frac{k g \cdot m}{s}}{45 k g} = 4 \frac{m}{s}$ $(180(kg*m)/s)/(45kg)= 4m/s$

Answer 2 · 2017-05-18T23:24:29

$4 m s^{- 1}$ $4ms^-1$

Explanation:

Momentum (written as $ρ$ $rho$ ) is defined as the product of the object's mass and it's velocity.

$ρ =$ $rho =$ mass x velocity
$180 = 45 v$ $180 = 45 v$
$v = 4 m s^{- 1}$ $v = 4 ms^-1$

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Question #ee704

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