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LekaFEV [45]
3 years ago
8

A spherically-spreading EM wave comes from a 104.0 W source. At a distance of 9.6 m, what is the intensity of the wave?

Physics
1 answer:
Nookie1986 [14]3 years ago
8 0

Answer:

Approximately 0.0898 W/m².

Explanation:

The intensity of light measures the power that the light delivers per unit area.

The source in this question delivers a constant power of \rm 104.0\; W. If the source here is a point source, that \rm 104.0\; W of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.

The surface area of a sphere of radius r is equal to 4\pi r^{2}. For the imaginary 9.6-meter sphere here, the surface area will be:

\rm 4\pi \times (9.6\; m)^{2} \approx 1158.12\; m^{2}.

That \rm 104.0\; W power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:

\displaystyle\rm  \frac{104.0\; W}{1158.12\; m^{2}}\approx 0.0898\; W\cdot m^{-2}.

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The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m
lions [1.4K]

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

8 0
3 years ago
An object is realsed From rest . how far does it fall during the second second of fall?​
Savatey [412]

Answer:

14.715 m

Explanation:

Assume that the acceleration due to gravity is 9.81 m/s^2 downwards, take downwards as positive

First second:

v = u + at

v = 9.81 m/s

Second second:

s = ut + (1/2)at^2

s = 9.81(1) + (1/2)(9.81)(1)^2

s = 14.715 m

6 0
3 years ago
What is the speed of a bobsled whose distance time graph indicated that it traveled 103m in 30s?
mars1129 [50]

v = s/t = (103 m) / (30 s) = 3.433 m/s = 12.36 km/h

3 0
3 years ago
What voltage produces a current of 50 amps with a resistance of 20 ohms?<br>​
PIT_PIT [208]

Answer:

1000 V

Explanation:

Ohm's law:

V = IR

V = (50 A) (20 Ω)

V = 1000 V

5 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid
ioda
<span>Radius = 4.6 m
 Time for one complete rotation t = 5.5 s.
 Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
 Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
 Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
  Force of the cat Fc = 6m, m being the mass.
 Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
 equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116</span>
4 0
3 years ago
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