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,


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 = 1336.67 sec = 22.27 Minutes.
So the maximum time will be 22 minutes 16 seconds.
Answer:
Explanation:
Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.
P = h d g
Where, h is the depth, d be the density and g be the acceleration due to gravity.
If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.
We can’t see the following
The intensity of a sound wave is defined as the amount of energy passing through a unit area of the wave front in unit of time.
Answer:
distance is 13 m for 100 dB
distance is 409 km for 10 dB
Explanation:
Given data
distance r = 2.30 m
source β = 115 dB
to find out
distance at sound level 100 dB and 10 dB
solution
first we calculate here power and intensity and with this power and intensity we will find distance
we know sound level β = 10 log(I/
) ......................a
put here value (I/
) = 10^−12 W/m² and β = 115
115 = 10 log(I/10^−12)
so
I = 0.316228 W/m²
and we know power = intensity × 4π r² ...............b
power = 0.316228 × 4π (2.30)²
power = 21.021604 W
we know at 100 dB intensity is 0.01 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 0.01 × 4π r²
so by solving r
r = 12.933855 m = 13 m
distance is 13 m
and
at 10 dB intensity is 1 × 10^–11 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 1 × 10^–11 × 4π r²
by solving r we get
r = 409004.412465 m = 409 km