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3 years ago

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A passenger at the rear of a train traveling at 15 m/s relative to earth throws a baseball with a speed of 15 m/s in the directi

on opposite the motion of the train. what is the velocity of the baseball relative to the earth as it leaves the thrower’s hand?

1 answer:

Natalija [7]3 years ago

4 0

Answer:

0 m/s

Explanation:

Let's take the direction in which the train is travelling as positive direction. Therefore:

- The velocity of the train with respect to earth is:

$v_t = +15 m/s$

- The ball is thrown with speed 15 m/s in the opposite direction, so in the frame of reference of the train, its velocity is

$v_b = -15 m/s$

Therefore, the velocity of the baseball in the earth's reference frame will be the algebraic sum of these two velocities, so:

$v_b' = v_t + v_b = +15 + (-15) = 0$

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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

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The object of badminton is to hit the birdie or "shuttlecock" over the net and onto to court to score how many points to win gam

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Answer:

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Explanation:

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A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

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Let $\omega$ be the angular velocity of Flywheel

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$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

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Answer:

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Answer:

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