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Firlakuza [10]
3 years ago
8

Consider a perfectly reflecting mirror oriented so that solar radiation of intensity i is incident upon, and perpendicular to, t

he reflective surface of the mirror. if the mirror has surface area a, what is frad, the magnitude of the average force due to the radiation pressure of the sunlight on the mirror
Physics
2 answers:
podryga [215]3 years ago
4 0
The answer is Frad<span> = 2IA/c.</span>
yKpoI14uk [10]3 years ago
4 0

To solve this problem we need to know the following:

the mass of the sun, Msun, 2.0 * 10^30 kg

the intensity of sunlight as a function of the distance R from the sun,

Isun(R) = 3.2 * 10^25 * 1/R2 (W/m2), and

the gravitational constant G = 6.67 * 10^-11 m3/(kg⋅s2)

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun.

What is the critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

Using the formula from Part A, find the force due to solar radiation:

Fr = 2IA/c

And compare to the force due to gravity:

Fg = G * m1 * m2 / R2

Fr = Fg

2IA/c = G * m1 * m2 / R2

Substitute in for I and the mass of the sun:

2 * (3.2 * 10^25 * 1/R2) * A / c = G * (2.0*10^30) * m2 / R2

The R2s cancel:

2 * (3.2 * 10^25) * A / c = G * (2.0 * 10^30) * m2

Now solve for m/A (where m = m2, the mass of the solar sail):

m/A = 2 * (3.2 * 10^25) / (c * G * (2.0*10^30))

m/A = 2 * (3.2 * 10^25) / ((3.0 * 10^8) * (6.67 * 10^-11) * (2.0 * 10^30))

mass/area = 0.00160 kg/m2

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Which month has the longest day and shortest night?
Volgvan
I think it’s June 20
8 0
3 years ago
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
Read 2 more answers
Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.
skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:  

a) The vectorial product, a×b  is:

a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k

b) The escalar product a·b is:

a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17

d) The component of a along the direction of b is:

a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66

I hope it helps you!                        

5 0
3 years ago
Plz Help me with this
AlexFokin [52]

Answer:

Cools ; size

Explanation:

The rate at which magma cools determines the size of the crystals in the new rock. Igneous rocks are formed from the cooling and solidification of molten magma which finds its way to the surface or depth of very low pressure beneath the surface. This place or depth of cooling of magma affects the cooling rate and hence the size of the crystals formed. Igneous rocks formed at depths below the surface have more time to cool and allows more time for Crystal growth and hence produce coarse grained crystal grains called Intrusive igneous rocks which have significantly larger crystals than those formed on the surface which cools rapidly and allowing very little time for crystal growth giving rise to the formation of fine grained crystals and are called extrusive igneous rocks.

6 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
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