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Tcecarenko [31]
3 years ago
12

A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total

area A of the panels is 10m2.1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?Express your answer numerically in kilowatts to two significant figures.2) What is the total force F on the panels exerted by radiation pressure from the sunlight?Express the total force numerically, to two significant figures, in units of newtons.
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

So lets say that the power absorbed

Will be = Intensity x Area

Which Is = 1.4 x 10^3 x(10)

So

14000 Watt = 14 kWatt

However we know that radiation pressure is equal to

time-averaged intensity all over the speed of light in free space

So

P = (1.4 x 1000)/c

But

F= P x A

So

((1.4 x 1000)/(3 x1 0^8)) x 10

Which is

=0.000046666N

To two SIG figures we have

=0.000047 N

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Nostrana [21]

Answer:

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Explanation:

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5 0
3 years ago
A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal
melamori03 [73]

Answer:

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

\sum M__A }=0

- 40 (180) - (80) F_E Cos 37° + 55  

-7200 + F_E (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79  F_E

- 30.79  F_E = 7200

F_E = -\frac{7200}{30.79}

F_E  = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

\sum f_y = 0

F_E sin 37 + F_A cos 20 - F_G = 0

- 233.84 sin 37 + F_A cos 20 - 180 = 0

F_A cos 20  = 320.73

F_A = \frac{320.73}{cos 20}

F_A = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

\sum f_y = 0

F_E cos 37 - F_{CB} + F_A sin 20° = 0

-233.84 cos 37 - F_{CB} + 341.31 sin 20° = 0  

-186.75 -  F_{CB} + 116.74 = 0

-  F_{CB} -70.01 = 0

-  F_{CB} = 70.01

F_{CB} = - 70.01 lb

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

7 0
4 years ago
Half Time At 3:00 the hour hand and the minute hand of a clock point in directions that are 90.0
jenyasd209 [6]

Answer:

3 : 08 : 10.9

Explanation:

assuming a 12 hour clock

angular velocity of the hour hand is 2π/(3600(12)) rad/s

angular velocity of the minute hand is 2π/3600 rad/s

difference is 2π/3600 - 2π/(3600(12)) = 11(2π/(3600(12)) rad/s

45° = π/4 radians

This angle is covered in a time of

π/4 rad / 22π / (3600(12)) = 900(12) / 22 = 490.909090... s

or about 8 minutes 10.9 s

ANSWER 3:08:10.9

5 0
3 years ago
A puddle dries up by the process of
levacccp [35]
The puddle dries up cause of the radiation from the sun
4 0
3 years ago
A tennis ball is dropped from 1.16 m above the
sweet-ann [11.9K]

Answer:

Vf = 4.77 m/s

Explanation:

During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 1.16 m

Vf = Final Velocity of Ball = ?

Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)

Therefore, using these values in the equation, we get:

(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²

Vf = √(22.736 m²/s²)

<u>Vf = 4.77 m/s</u>

6 0
3 years ago
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