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Vedmedyk [2.9K]
3 years ago
10

A 20-kg cart travels at 20 m/s in a circle with a radius of 20 m. What is the centripetal force?

Physics
2 answers:
tia_tia [17]3 years ago
8 0

Answer:

The equation for centripetal force is:

T = (mv^2)/r

T = centripetal force

m = mass

v = speed

r = radius

T = [20kg * (20 m/s * 20m/s)] / 20 m

T = [20kg * 400 m/s^2] / 20 m

T = 8,000 kg m/s^2 / 20 m

T = 400 N

Explanation:

Vlad1618 [11]3 years ago
5 0

Centripetal force = (mass) x (speed)² / (radius)

                           = (20 kg) x (20 m/s)² / (20 m)

                           = (20 x 20 / 20)  (kg-meter/sec²)

                           = 20 newtons


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Adhesive.

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Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is
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Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity g of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

F=G\frac{m_{E}m}{r^2}  (1)

Where:

F is the module of the force exerted between both bodies

G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}} is the gravitational constant

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R_{E}=6.38(10)^{6} m is the radius of the Earth

h=3.59(10)^{7} m is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

F=mg  (2)

Combining (1) and (2):

G\frac{m_{E}m}{r^2}=mg  (3)

Isolating g:

g=\frac{G M_{E}}{r^2}  (4)

Remembering r=R_{E}+h:

g=\frac{G M_{E}}{(R_{E}+h)^2}  (5)

g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}  

Finally:

g=0.223 m/s^{2}  

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3 years ago
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