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Elan Coil [88]
3 years ago
10

The intensity of sunlight falling on the earth is about 1.4 kw/m2 (before any gets absorbed by our atmosphere). at what rate doe

s the sun emit light energy? (the earth-sun distance = 1.5 × 108 km and the earth's radius = 6.4 × 103 km.)
Physics
1 answer:
xxTIMURxx [149]3 years ago
6 0
The area of the Earth (Ae) that is irradiated by is given by:

Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2

On the Earth, insolation (We) = Psun/Ae

Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W
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Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

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3 years ago
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Answer:

550 kg

Explanation:

mass = E / gh

= 33000/60

=550

plzzz......

mark it as a brilliant answer

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Answer:

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