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Mkey [24]
3 years ago
7

9) Of all the types of light the Sun gives off, it emits the greatest amount of light at visible wavelengths of light. If the Su

n were to cool off dramatically and as a result start giving off mainly light at wavelengths longer than visible light, how would the frequency, energy, and speed of this light given off by the Sun also be different? Explain your reasoning.
Physics
1 answer:
erastova [34]3 years ago
7 0

Answer:

*  most of the emission would be in the infrared part, the visible radiation would be very small.

*total intensity of the semition decreases that the intensity depends on the fourth power of the temperature

Explanation:

The radiation emitted by the Sun is approximately the radiation of a black body, if the Sun were to cool, the maximum emission wavelength changes

          λ T = 2,898 10⁻³

          λ = 2,898 10⁻³ / T

if the temperature decreases the maximum wavelength the greater values ​​are moved, that is to say towards the infrared. Therefore the emission curve also moves, in this case most of the emission would be in the infrared part, the visible radiation would be very small.

Furthermore, the total intensity of the semition decreases that the intensity depends on the fourth power of the temperature according to Stefan's law

           P = σ A eT⁴

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Answer:

Inertia depends on mass and density.

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3 years ago
A) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a
MrMuchimi

a) 1.96\cdot 10^{-16} m/s^2

The gravitational field strength near the surface of the asteroid is given by:

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the asteroid

R the radius of the asteroid

h is the distance from the surface

Substituting the data of the asteroid:

M=3.2\cdot 10^3 kg is the mass

R=30 km = 30000 m is the radius of the asteroid

h=3 km = 3000 m is the distance from the surface

We find

g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2

b) i)  5.53\cdot 10^9 s

The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

g=1.96\cdot 10^{-16} m/s^2

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

s=ut+\frac{1}{2}gt^2

where

s = 3 km = 3000 m is his displacement to reach the surface

u = 0 is his initial velocity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s

b) ii) 1.08\cdot 10^{-6} m/s

Again, we can use another suvat equation:

v=u+gt

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

Since we have

u = 0

t=5.53\cdot 10^9 s

g=1.96\cdot 10^{-16} m/s^2

The velocity of the astronaut at the surface will be

v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s

b) iii) 175 years

The duration of one year here is

T=3.16\cdot 10^7 s

And the time it takes for the astronaut to reach the surface of the asteroid is

t=5.53\cdot 10^9 s

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175

So, the astronaut will take 175 years to reach the surface.

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3 years ago
An elephant pushes with 2000 N on a load of trees. It then pushes these trees for 150 m. How much work did the elephant d
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Answer:

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I believe it is relative humidity.
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In the question, you just gave a complete and detailed
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You might want to simplify the speed and make it a unit rate,
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6 0
3 years ago
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