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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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Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an
VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given \Delta x=7\ m

\theta=45 \°

Also, \Delta y=(3.5-2)=1.5\ m

a_x=0\ and\ a_y=-9.81\ m/s^2

Let us say the velocity in the x-direction is v_x and in the y-direction is v_y. And acceleration in the x-direction is a_x and in the y-direction is a_y.

Also, \Delta x\ and\ \Delta y is distance covered in x and y direction respectively. And t is the time taken by the ball to hit the backboard.

We can write v_x=v_0cos(45)\ and\ v_y=v_0sin(45). Where v_0 is velocity of ball.

Now,

\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt

\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}

Also,

\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s.

Plugging this value in

t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}

t=1.06\ seconds

So, the time of flight of the ball is 1.06 seconds.

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