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Oliga [24]
4 years ago
12

A satellite is in a circular orbit around the earth. the period of the satellite is 25.4 hr. calculate the radius of the orbit o

f the satellite. data: mass of the earth = 5.98 x 1024 kg.
Physics
1 answer:
azamat4 years ago
8 0
Satellite X has a greater period an a slower tangential speed than satellite Y
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tekilochka [14]
The integral of acceleration is velocity. The area under the curve is an integral. You can see the relation here. So just take the area under those time intervals. I’m assuming 1.7 is where it crosses the x-axis

For the first interval:
Base = 1.7
Height = -2
[1.7*(-2)]/2 = -1.7 cm/s

For the second interval:
There are two triangles and a rectangle here.
Base #1 = 0.3
Height #1 = 1

Base #2 = 1
Height #2 = 1

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Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration
Digiron [165]

Answer:

Approximately 2.05\; {\rm m\cdot s^{-2}}.

Explanation:

The net force on the girl would be:

\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}.

Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.

In other words, the tension force that the rope exerted on the girl would be 123.0\; {\rm N}. The girl would exert a reaction force on the rope at the same magnitude (123.0\; {\rm N}\!) in the opposite direction. This force would translate to a 123.0\; {\rm N}\!\! force on the boy towards the girl.

Under similar assumptions, the net force on the boy would also be 123.0\; {\rm N}. Since the mass of the boy is m(\text{boy}) = 60\; {\rm kg}, the acceleration of the boy would be:

\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}.

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Alexxandr [17]
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goblinko [34]
D. all of these

hope this helps.
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4 years ago
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