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Free_Kalibri [48]
3 years ago
15

A 2.0-kg object moving at 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy l

ost by the system as a result of this collision.
Physics
1 answer:
jekas [21]3 years ago
7 0

Answer:

20 J

Explanation:

From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.

m'u'+mu = V(m+m')........... Equation 1

Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.

make V the subject of the equation above

V = (m'u'+mu)/(m+m')............. Equation 2

Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).

Substitute into equation 2

V = [(2×5)+(8×0)]/(2+8)

V = 10/10

V = 1 m/s.

Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision

Total kinetic energy before collision = 1/2(2)(5²) = 25 J

Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J

Lost in kinetic energy = 25-5 = 20 J

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Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
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Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

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The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

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You have not provided the diagram, therefore, I cannot provide an exact answer.
However, I will try to help by explaining how to solve this problem.

When light moves from air to glass:
1- part of the light is reflected back into the air where the angle of incidence is equal to the angle of reflection
2- part of the light enters the water and refracts. The angle of refraction can be calculated using Snell's law.

In a diagram, the reflected ray would be the one getting back into air while the refracted ray would be the one entering the water.

You can check the attached diagram for further illustrations.

Hope this helps :)

6 0
3 years ago
Read 2 more answers
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