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Citrus2011 [14]
3 years ago
11

The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of

the sun. (Take e = 1, and σ = 5.67 × 10−8W/m2⋅K4).
a. 3.95 × 1026 W
b. 5.17 × 1027 W
c. 9.62 × 1028 W
d. 6.96 × 1030 W
Physics
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

a. 3.95\times10^{26}W

Explanation:

T = temperature of the surface of sun = 5800 K

r = Radius of the Sun = 7 x 10⁸ m

A = Surface area of the Sun

Surface area of the sun is given as

A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}

e = Emissivity = 1

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the sun is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W

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3 years ago
a tennis ball is thrown straight up at a speed of 40m/s and caught at the same level. calculate rhe maximum height reached by th
Rudiy27

Answer:

81.6 m

Explanation:

Answer: 81.6 m.

The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is

-v(initial) / -g = t

-40 m/s / -9.8 m/s^2 = 4.08 s

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7 0
3 years ago
Read 2 more answers
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe
Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A

Therefore, the bottom current is 12.8 A to the right.

3 0
2 years ago
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Work = (weight) · (height) = (50kg) · (9.8 m/s²) · (6 m)

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Power = (50 · 9.8 · 6 / 15) · (kg · m² / s³)

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Power = 196 Newton-meter/second

<em>Power = 196 watts</em>

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Answer:

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