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3 years ago

What is the mass of a child in a wagon that has a velocity of 10 m/s and a Momentum of 30 KG* M/S

Physics

1 answer:

LenKa [72]3 years ago

3 0

Explanation:

sinces : Momentum = velocity × mass

then : 30 = 10 × m and m = 30 ÷ 10 = 3 kg

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A ball is moving in a certain direction. What could happen to the ball if a greater force was applied on the ball along its dire

alukav5142 [94]

Answer:

the one going faster would prolly stop and the one it hit would start rolling the opposite direction it was. like think about if u were playing pool.

Explanation:

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2 years ago

Which of the following is described here? "Point your skis straight down the fall line with your skis parallel and about a foot

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answer d. down hill schussing

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3 years ago

Read 2 more answers

Two automobiles of equal mass approach an intersection. One vehicle is traveling with speed 13.0 m/s toward the east, and the ot

puteri [66]

Answer:

Explanation:

We shall apply law of conservation of momentum to know the Speed of northward moving vehicle before collision to check the veracity of driver's statement .

Let v be the velocity of composite mass after collision

Applying law of conservation of momentum in north direction

m v₂ = 2m v sin55.08

Applying law of conservation of momentum in east direction

m x 13 = 2m v cos55.08

Dividing these two equations

v₂ / 13 = tan55.08

v₂ = 13 tan55.08

= 18.62 m/s

= (18.62 x60 x 60) / 1000

= 67 km/h

= 67 x 5/8 mi/h

= 42 mi/h

So he is lieing.

7 0

4 years ago

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial: