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kkurt [141]
3 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]3 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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The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N. 

<span>Fx = [(233 + 840)/g]*v²/7.5 </span>

<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>

<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>

<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>

<span>233 + 840 = Ti*cos40º </span>

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Read 2 more answers
In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.
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The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

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given information:

s = 220 m

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he unrealistically large acceleration experienced by the space travelers during their launch

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a = 2.79 x 10⁴ g

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