2 years ago

Two pucks collide with each other. after the collision they do not stick together. is kinetic energy conserved in the collision?

1 answer:

5 0

Since they do not stick after collision hence collision is elastic. In elastic collision, both momentum and kinetic energy is conserved because in this type of collision, first body deforms but then quickly regains its former shape and transfers its kinetic energy to the second pluck.
So kinetic energy is conserved.

You might be interested in

An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of

Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

$3\times \frac{\lambda}{2} = 0.39$

$\lambda = 0.26 m$

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

$d = \frac{\lambda}{4}$

$d = \frac{0.26}{4} = 0.065 m = 6.5 cm$

So the distance will be 6.5 cm

3 0

2 years ago

A motorist driving at 25 meters/second decelerates to

Ilia_Sergeevich [38]

Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.

Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,

Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules

Just ignore the negative value for the final result because work is a scalar quantity.

8 0

3 years ago

Read 2 more answers

Nalah labels ten rats with the numbers 1 through 10. Then, she records how long it takes each rat to run through a maze. She fin

drek231 [11]

Answer:

we have to make another question we maxed it out again lol

Explanation:

5 0

3 years ago

A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a

valkas [14]

The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
$\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}$

The rate of change of the radius is given as
$\frac{dr}{dt} =-2 \, \frac{ft}{s}$
Therefore
$\frac{dA}{dt} =-4 \pi r \, \frac{ft^{2}}{s}$

When r = 10 ft, obtain
$\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi \, \frac{ft^{2}}{s}$

Answer: - 40π ft²/s (or - 127.5 ft²/s)

7 0

3 years ago

Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight

svetlana [45]

I., II., and IV. are examples of acceleration. III. isn't.

6 0

3 years ago