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love history [14]
3 years ago
5

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?
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Physics
1 answer:
koban [17]3 years ago
7 0

Answer:

0.00004666N

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

Explanation:

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The figure shows two springs (k1 = 10 N/m and k2 = 20 N/m ) attached to a block that can slide on a frictionless surface. In the
Rainbow [258]

Answer:

Explanation:

a )

Energy stored by left spring when compressed = 1/2 k x²

= .5 x 10 x .02² = .002 J .

Let compression in right spring = y

energy stored to right spring = 1/2 k y²

1/2 k y² = 0.002

.5 x 20 x y² = 0.002

y = .01414 m

= 1.4  cm

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3 years ago
Which of these was Sir Isaac Newton not known for? invention of calculus invention of the microscope the Three Laws of Motion un
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3 0
3 years ago
Read 2 more answers
Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b
Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}

And also,

R^{2}=R^{2} _{0}A^{\frac{2}{3} }

And, R _{0}=1.2\times 10^{-15}m

Now,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0
3 years ago
A model plane has a mass of 0.75 kg and is flying 12 m above the ground
Grace [21]

Answer:

Option C. 210 J.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Velocity (v) = 18 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Total Mechanical energy (ME) =?

Next, we shall determine the potential energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Acceleration due to gravity (g) = 9.8 m/s²

Potential energy (PE) =?

PE = mgh

PE = 0.75 × 9.8 × 12

PE = 88.2 J

Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Velocity (v) = 18 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.75 × 18²

KE = ½ × 0.75 × 324

KE = 121.5 J

Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:

Potential energy (PE) = 88.2 J

Kinetic energy (KE) = 121.5 J

Total Mechanical energy (ME) =?

ME = PE + KE

ME = 88.2 + 121.5

ME = 209.7 J

ME ≈ 210 J

Therefore, the total mechanical energy of the plane is 210 J.

8 0
2 years ago
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