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Orlov [11]
3 years ago
8

A small model car with mass m travels at a constant speed on the inside of a track that is a vertical circle radius 5.00 m. If t

he normal force exerted by the track on the car when it is at the bottom of the track (point A) is equal to 2.50 mg, how much time does it take the car to complete one revolution around the track?
Physics
1 answer:
dolphi86 [110]3 years ago
6 0
One point to another
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A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
erastova [34]
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
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v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
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45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

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A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'
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Explanation:

6 0
2 years ago
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A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte
musickatia [10]

Answer:

Number of electrons, n=5.62\times 10^{21}

Explanation:

It is given that,

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Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, I=\dfrac{q}{t}

I\times t=n\times e

n=\dfrac{It}{e}

e is the charge of an electron

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n=5.62\times 10^{21}

So, the number of electrons pass through the resistor is 5.62\times 10^{21}. Hence, this is the required solution.

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