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Phantasy [73]
3 years ago
14

During its final days as a red giant, the Sun will reach a peak luminosity of about 3000LSun. Earth will therefore absorb about

3000 times as much solar energy as it does now, and it will need to radiate 3000 times as much thermal energy to keep its surface temperature in balance. Estimate the temperature Earth’s surface will need to attain in order to radiate that much thermal energy.
Physics
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

2124.03764889 K

Explanation:

Luminosity of Sun = L

New luminosity of Sun = 3000L

T_1 = Current temperature of Earth = 287 K (assumed)

Luminosity is given by

L=4\pi r^2\sigma T^4

Here,

L\propto T

\dfrac{3000L}{L}=\dfrac{T_2^4}{T_1^4}\\\Rightarrow 3000=\dfrac{T_2^4}{T_1^4}\\\Rightarrow 3000^{\dfrac{1}{4}}=\dfrac{T_2}{T_1}\\\Rightarrow T_2=3000^{\dfrac{1}{4}}\times 287\\\Rightarrow T_2=2124.03764889\ K

The temperature of the Earth's surface is 2124.03764889 K

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The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a
jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A V_a=382\ V

Potential at V_c=785\ V

when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1

Change in Kinetic Energy of particle moving under the Potential From C to B

q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2

Divide 1 and 2 we get

\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}

on solving we get

V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c

V_b=\frac{743}{3}=247.67\ V

                     

4 0
3 years ago
What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency
Anvisha [2.4K]

Answer:

1.04\times 10^{-7} T

Explanation:

IP  = Power of the bulb = 100 W

r  = distance from the bulb = 2.5 m

I = Intensity of light at the location

Intensity of the light at the location is given as

I = \frac{P}{4\pi r^{2}}

I = \frac{100}{4(3.14) (2.5)^{2}}

I = 1.28 W/m²

B_{o} = maximum magnetic field

Intensity is given as

I = \frac{B_{o}^{2}c}{2\mu _{o}}

1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}

B_{o} = 1.04\times 10^{-7} T

7 0
3 years ago
A person is attracted towards the center of the earth by an 800 N gravitational force. The force with which the earth is attract
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Answer:

800 N

Explanation:

By Newton's third law which states that for every action, there is an equal and opposite reaction.

So, as the earth attracts the person towards its center, the person attracts the earth towards itself with the same magnitude of force but in the opposite direction.

Since the person is attracted towards the center of the earth by an 800 N gravitational force, the  the earth is attracted toward the person with an 800 N reaction force.

7 0
2 years ago
A circle graph shows that one-fourth of the students in a particular school ride the bus every day. What is this equivalent meas
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Answer:

it's 1.57... radians and 90°

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3 years ago
As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th
Lena [83]

Answer:

Pressure will be 6072449.952Pa

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

P=\frac{F}{A}=\frac{mg}{A}

So force F = mg = 65×9.8 = 637 N

Area A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2

So pressure p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa

5 0
3 years ago
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