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svlad2 [7]
2 years ago
15

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

Physics
1 answer:
Hitman42 [59]2 years ago
7 0

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

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Sliva [168]
Power = I^2 x R
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4 0
3 years ago
A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be
MrRissso [65]

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0
3 years ago
Brainly app · Installed
Sauron [17]
I need a picture plz I don’t know what to answer.
6 0
3 years ago
26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the
denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

U=mg \Delta h

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h = 4 m + 0.02 m = 4.02 m is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J

4 0
2 years ago
The repeal of the Trustee's ban on slavery boosted Georgia's economy by
Leviafan [203]
The right answer for the question that is being asked and shown above is that: "B) allowing Carolina planters to expand rice cultivation into Georgia's lowlands."
-I hope this is the answer you are looking for.


8 0
3 years ago
Read 2 more answers
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