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mamaluj [8]
3 years ago
14

As the object in the picture moves from point A to point D across the surface, the sum of the gravitational and kinetic energies

_____. A. decreases only B. increases and then decreases. C. decreases and then increases. D. remains the same. The object will have the maximum Kinetic Energy at point___. A B C D The object will have the maximum Potential Energy at point____. A B C D If the car has a mass of 1000kg and starts at rest, what is the speed of the car at point C
Physics
1 answer:
Afina-wow [57]3 years ago
7 0
Answer:
d. remains the same
Explanation:
We know that if there is no external resistive force then the total energy of the system remains constant.
The total energy is the summation of kinetic and potential energy.
When an object moves from point A to another point D then the total energy, potential gravitational and kinetic energy will remain same.

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A car travels 100 m while decelerating to 8 m/s in 5 s.<br> a) What was its initial speed?
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Answer:

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

v_{f} =v_{i} - (a*t)\\

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

Vf = final velocity = 8[m/s]

Vi = initial velocity [m/s]

a = acceleration = [m/s^2]

t = time = 5 [s]

Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

v_{f}^{2} = v_{i}^{2} - (2*a*d)

where:

d = distance = 100[m]

(8^2) = (8 + 5*a)^2 - (2*a*100)

64 = (64 + 80*a + 25*a^2) - 200*a

0 = 80*a - 200*a + 25*a^2

0 = - 120*a + 25*a^2

0 = 25*a(a - 4.8)

therefore:

a = 0 or a = 4.8 [m/s^2]

We choose the value of 4.8 as the acceleration value, since the zero value would not apply.

Returning to the first equation:

8 = Vi - (4.8*5)

Vi = 32 [m/s]

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