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enyata [817]
3 years ago
15

A satellite is placed in a geosynchronous orbit. In this equatorial orbit with a period of 24 hours, the satellite hovers over o

ne point on the equator. Which statement is true for a satellite in such an orbit?a. There is a tangential force that helps the satellite keep up with the rotation of the Earth.b. There is no acceleration toward the center of the Earth.c. The force toward the center of the Earth is balanced by a force away from the center of the Earth.d. The satellite is in a state of free fall toward the Earth.e. There is no gravitational force on the satellite.
Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

see that the correct one is D

Explanation:

In this problem the satellite is subjected to the universal force of attraction

    F = G M m / r²

using Newton's second law with centripetal acceleration you can find the angular velocity of the satellite

  F = m a

  F = m w² r

  G M / r³ = w²

and   w = 2π / T

In this case, the satellite is in a two-dimensional movement, where the free fall of the satellite is compensated by the horizontal displacement, for which it is always at a distance from the earth, remaining at the same point.

When reviewing the different statements we see that the correct one isD

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A motorcycle is moving at a constant speed of 40km/h. How long does it take the motorcycle to travel a distance of 10km?
nlexa [21]
Time = distance/time
time(hours)= 10/40
time(hours)=0.25
=15 minutes
7 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
3 years ago
Heterotrophs convert solar energy into chemical energy.<br> a. True<br> b. False
jonny [76]

The answer would be false

3 0
4 years ago
Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript
Pavel [41]

Answer:

Part a)

F_A = 4.59 N

Part B)

F_B = 1.28 N

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

F_A + F_B = ma

as we know that

a = 0.554 m/s^2

m = 10.6 kg

now we will have

F_A + F_B = 10.6(0.554)

F_A + F_B = 5.87 N

Now two forces are in opposite direction then we have

F_A - F_B = 10.6(0.313)

F_A - F_B = 3.32 N

Part A)

Now we will have from above two equation

F_A = 4.59 N

Part B)

Similarly for other force we have

F_B = 1.28 N

5 0
3 years ago
A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude
kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

v_0=115 m/s (toward east)

So, decomposing along the x- and y- directions:

v_{x0} = 115 m/s\\v_{y0} = 0

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

a=2.88 m/s^2 (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2

So the two components of the velocity after a time t = 25.0 s will be

v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s

So, the magnitude of the velocity of the plane will be

v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s

6 0
3 years ago
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