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kkurt [141]
2 years ago
7

The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear

th's axis of rotation is tilted by 27° relative to the plane of the orbit (see the drawing), so sunlight does not strike the equator perpendicularly. What power strikes a 0.570-m2 patch of flat land at the equator at point Q?
Physics
1 answer:
WITCHER [35]2 years ago
6 0

Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

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using the law of the conservation of the linear momentum:

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where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

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270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

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\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

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\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

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so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

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