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dusya [7]
3 years ago
10

A 0.250 kg car rest rolls down a frictionless incline from a starting height of 0.75 m what is the final velocity of the car? SH

OW ALL WORK.
Also what difference would it make if it is not frictionless? Please and thank you!
Physics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

v = 3.83 m/s

Explanation:

Given that,

The mass of a car, m = 0.250

It rolls down a frictionless incline from a starting height of 0.75 m

We need to find the final velocity of the car. Let v be the velocity of the car. It can be calculated using the third equation of motion i.e.

v^2-u^2=2as

Here,

u = 0 and a = g

v=\sqrt{2gs} \\\\v=\sqrt{2\times 9.8\times 0.75} \\\\v=3.83\ m/s

So, the final velocity of the car is 3.83 m/s.

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Answer:

W=\frac{773}{4.45}=173.76 l b f

Explanation:

W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}

The law of gravitation

G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)

Universal gravitational constant [S.I. units]

m_{e}=5.976\left(10^{24}\right) k g

Mass of Earth [S.I. units]

m=89 kg

Mass of a man in a spacecraft [S.I. units]

R=6371 \mathrm{~km}

Earth radius [km]

Distance between man and the earth's surface

h=261 \mathrm{~km} \quad[\mathrm{~km}]

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W=\frac{773}{4.45}=173.76 l b f

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2 years ago
I NEED THIS A SOON AS POSSIBLE
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Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

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In a series RLC ac circuit, a second resistor is connected in series with the resistor previously in the circuit. As a result of
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Answer:

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