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3 years ago

What is the velocity of a car with a momentum of 100 kg*m/s and a mass of 35kg?

Physics

1 answer:

LuckyWell [14K]3 years ago

4 0

Answer:

Explanation:

The velocity of the car can be found by using the formula

$v = \frac{p}{m} \\$

p is the momentum

m is the mass

From the question we have

$v = \frac{100}{35} = \frac{50}{7} \\ = 2.857142...$

We have the final answer as

Hope this helps you

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A pitcher throws a 0.14-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

kykrilka [37]

Answer

given,

mass of base ball = 0.14 kg

speed before it made the contact with the ball (V i) = 42 m/s

speed after batter hit the ball(V f) = - 48 m/s

impulse = change in momentum

= $m\times (V_f-V_i)$

= $0.14\times (-48-42)$

= -12.6 Kg m/s

Magnitude of impulse = 12.6 Kg m/s

Force = $\dfrac{impulse}{time}$

= $\dfrac{12.6}{0.005}$

Force = 2520 N

5 0

3 years ago

Plz answer these two questions. I am so sick.if u ddo it would mean a lot.

faltersainse [42]

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2 years ago

A rock falls 15 m. is this vertical or horizontal motion? what is the displacement?

Irina-Kira [14]

Displacement will be 15 too because
It falls from 0 till 15 meters

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3 years ago

What is the average speed of a car that travels 60 meters in 2<br> seconds?

blondinia [14]

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

6 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago