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3 years ago

MathPhys May you help me once again?

Physics

1 answer:

Maru [420]3 years ago

6 0

Answer:

2. Same force; same impulse; Mini Cooper has greater change in velocity

4. Same momentum

6. It is distributed to the two halves

Explanation:

I will do the even problems as examples.

2. From Newton's third law, we know the bulldozer and Mini Cooper will experience equal but opposite forces. So the magnitude of the force is the same.

Impulse is force times time. Since the force is the same, and the amount of time is the same, then the impulse is the same.

Impulse is equal to change in momentum, or mass times change in velocity. Since the Mini Cooper has the smaller mass, it has the greater change in velocity.

4. Momentum is mass times velocity. The weight of the rover is different on Mars than on Earth, but the mass is still the same. Therefore, if the rover moves at the same speed, it will have the same momentum.

6. The pencil's momentum is conserved. When the pencil breaks in half, half the momentum goes to one half of the pencil, and half the momentum goes to the other half of the pencil.

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Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

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erastovalidia [21]

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