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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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Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the
Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) ,   is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0
3 years ago
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from t
mestny [16]

Answer:

Wavelength is 471 nm

Explanation:

Given that,

Lines per unit length of diffraction grating is 500 lines/mm.

The third maxima from the central maxima (m=3) is at an angle of 45°

We need to find the color of laser light shines through a diffraction grating.

The condition for maxima is :

d\sin\theta=m\lambda

d = 1/N, N = number of lines per mm

\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm

6 0
3 years ago
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