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Orlov [11]
4 years ago
9

1. A ball bouncing against the ground and rebounding is an example of an elastic collision. Describe two different methods of ev

aluating this interaction, one for which momentum is conserved, and one for which momentum is not conserved. Explain your answer.
Physics
1 answer:
Rudik [331]4 years ago
4 0

Answer:

One method to solve this problem is to take the ball and the floor as the system

second method to solve this exercise and we take the system as formed by the ball only

Explanation:

One method to solve this problem is to take the ball and the floor as the system, therefore the forces that occur during the collision are driving force and reaction, internal to the system, therefore the moment is conserved.

A second method to solve this exercise and we take the system as formed by the ball only, in this case there is a change at the moment and this is not preserved, This second case is a little more difficult to solve

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If a certain mass has its velocity changed from 6.00 m/s to 7.50 m/s when a 3.00 n force acts for 4.00 seconds, find the mass of
shutvik [7]
By definition we have that the force for time is equal to the product of the mass for the change in speed.
 We have then that
 F * (delta t) = m * (delta v)
 Clearing the mass
 m = (F * (delta t)) / (delta v)
 Substituting the values
 m = ((3.00) * (4.00)) / (7.50-6.00) = 8
 answer
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3 0
4 years ago
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How many seconds did it take (after starting his descent) for the worker to hit the ground? Answer in units of s.
elena-14-01-66 [18.8K]

Answer:

This question is incomplete

Explanation:

The question is incomplete. However, to determine the time (in seconds) it took a worker to hit the ground from an elevated point. The speed the worker was coming with to the ground and the distance between the elevated point and the ground will have to be considered. Thus the formula to be used here will be

Speed (in meter per second) = distance (in meters) ÷ time (in seconds)

time (in seconds) = distance (in meters) ÷ speed (in meter per seconds)

6 0
3 years ago
What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpen
prisoha [69]

Answer:

The moment of inertia I is

I = 2.205x10^-4 kg/m^2

Explanation:

Given mass m = 0.5 kg

And side lenght = 0.03 m

Moment of inertia I = mass x radius of rotation squared

I = mr^2

In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.

Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m

Therefore,

I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2

8 0
4 years ago
PLEASE ANSWERRRRR ASAPPPPPP
Bad White [126]
D Valence
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5 0
3 years ago
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The capacity of the air to hold water vapor: Group of answer choices 1. decreases with an increase in temperature.2. increases w
Nezavi [6.7K]

Answer:

3. increases with an increase in temperature.

Explanation:

The air more water vapor at higher temperatures because at higher temperatures the air expands and the inter-molecular space increases so the room for water molecules increases.

Warm air keeps the water molecules warm and prevents them from condensing.

The air can hold the moisture only upto its saturation quantity after which the precipitation occurs in the form of rain, snow, hail, sleet etc.

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