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Lapatulllka [165]
2 years ago
8

A satellite is orbiting Earth in an approximately circular path. It completes one revolution each day (86,400 seconds). Its orbi

tal distance is 22,500 kilometers from the Earth's center. What is the satellite's average speed?
a. 5.5 kilometers/second
b. 19.4 kilometers/second
c. 1.64 kilometers/second
d. 384 meters/second
Physics
1 answer:
Vaselesa [24]2 years ago
8 0

Circular path ... circumference = 2π · Radius

Radius = distance from the Earth's center = 22,500 km

Circumference = 2π · 22,500 km = 141,372 km

Speed = (distance to be covered) / (time to cover the distance)

Speed = (141,372 km) / (86,400 seconds)

Speed = (141,372/86,400) (km/sec)

<em>Speed = 1.64 km/sec</em>


<em>Extra Info:</em>

This problem is a nice exercise in Arithmetic, but you shouldn't take any of it to run with in the real world.

A satellite with a period of 1 day is NOT  22,500 km from the center of the Earth.  It's more like 22,500 <u>MILES </u>from the Earth's <u>SURFACE. </u>

The real number is 22,236 miles above sea level, and 26,199 miles (42,164 km) from the center of the Earth.

To do <u>THAT</u> orbit in 86,400 seconds, the satellite has to zip around at <em>3.07 km/sec</em>.

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Answer:

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We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

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We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

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Thus, to get the average velocity for the first half of the swim

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Average velocity for the second half of the swim will be calculated in like manner, thus,

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A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
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Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

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Q is the charge on the sphere

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Here we have:

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We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

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Since the voltage is inversely proportional to r, we can use:

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