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vlada-n [284]
3 years ago
8

The centripetal force acting on a satellite in orbit

Physics
2 answers:
krok68 [10]3 years ago
8 0

The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.

AysviL [449]3 years ago
3 0

Answer : (D) " all of the above "

Explanation :

The forces acting on a satellite are Earth's gravitational force and centripetal force. The properties of centripetal force acting on a satellite in orbit are as follows :

(1) It acts as an unbalanced force on the satellite. This is due to the reason that there is no same force present which pulls the satellite away from the earth.

(2) It changes the direction of the satellite. The satellite is revolving with a constant speed while the direction is kept on changing (in a circular motion).

(3) Centripetal force always acts towards the center.

So, the correct option is (D) " all of the above "

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A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un
KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

x=vt=23.3\frac{m}{s}t

and the police car distance:

x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}

Since they both travel the same distance x, we can equal both formulas and solve for t:

0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\

Two solutions exist to the equation; the first one being t=0

The second solution will be:

0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s

This result allows us to confirm that the police car will take 32s to catch up to the speeder

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This question is incomplete because the options are missing; here is the complete question

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