If the total time is 120 seconds and the total distance is 320 meters. Calculate

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0

2 years ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

3 years ago

Read 2 more answers

A book is sliding across a horizontal desk and comes to a stop. the books kinetic energy was converted into what type of energy

Oliga [24]

Friction stole the book's kinetic energy, and turned it into heat energy ... which blew away in the breeze.

4 0

3 years ago

If you have 100 W expended over 20 s how much energy did it take?

netineya [11]

Power = (work or energy) / (time)

100 W = (energy) / (20 sec)

Energy = 2,000 watt-sec

<em>Energy = 2,000 J</em>

8 0

2 years ago

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago