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Naya [18.7K]
3 years ago
8

how long would it take for a radio wave sent from a space satellite circling mars to reach Earth? Assume that radio waves (a for

m of electromagnetic radiation) travel at the speed of light?
Physics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

3 Minutes 2 seconds to 22 minutes 16 seconds

Explanation:

Lets assume the Mars to be at the closest distance to Earth. This distance (D) = 54.6 Million km

The signal travels at the speed of light, so speed of the signal (V) = 300000 km/s

So, the time (T) taken by the radio wave to reach Earth from Mars will be,

T = \frac{D}{V}

T = \frac{54600000}{300000}

Thus, T = 182 Sec = 3 Minutes 2 seconds.

The radio wave will take minimum 03 Minutes 02 seconds to reach Earth. Here is should be noted that the distance between the two planets keep on changing as they revolve around the Sun. There will come a point when Mars is farthest from Earth and the distance (D) will be 401 Million km. Then, the time will change to ,

T = \frac{401000000}{300000}

T = 1336.67 sec = 22.27 Minutes.

So the maximum time will be 22 minutes 16 seconds.

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Explanation:

Input power  = 220(5) = 1100 W

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Neglecting hysteresis and other minor losses, the power will remain the same.

4 0
2 years ago
If a body accelerates from a state of relative rest in a gravity field, where does the energy come from to transfer as kinetic e
Delvig [45]

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5 0
2 years ago
Meteorologists classify tropical cyclones using a measurement of sustained winds within the low-pressure system. How do meteorol
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Explanation:

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3 0
3 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

4 0
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Nata [24]

Answer:

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Explanation:

just search it up

8 0
3 years ago
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