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Ahat [919]
3 years ago
9

What did Edwin Hubble discover about galaxies that allowed him to determine the expansion rate of the universe?

Physics
1 answer:
mario62 [17]3 years ago
6 0
<span> The farther away a galaxy is, the more its light
is shifted toward the red end of the spectrum. (C)</span>
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Find the intensity of electromagnetic radiation at the surface of the sun (radius r=R=6.96×105kmr=R=6.96×105km). Ignore any scat
alisha [4.7K]

Answer:

I = 4.46*10^{16}W/m^2.

Explanation:

Intensity I of the electromagnetic radiation is given by

I = \dfrac{P}{4\pi r^2},

where r is the distance from the EM source (the center of the sun, in our case), and P is the power output of the sun and it has the value

P = 3.9 *10^{26}W.

Since the radius of the sun in meters is r = 6.96*10^8km, the intensity I of the electromagnetic radiation at the surface of the sun is

I = \dfrac{3.9*10^{26}W}{4\pi (6.96*10^8m)^2}\\\\\boxed{ I = 4.46*10^{16}W/m^2}

The intensity of the electromagnetic radiation at the surface of the sun is I = 4.46*10^{16}W/m^2.

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3 years ago
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Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC
Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

q_1=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 1)

q_2=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N

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3 years ago
What is the value of acceleration as a body moves with an uniform speed
Levart [38]

Answer:

the value of acceleration as a body moves with an uniform speed is zero

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2 years ago
only a relatively small part of the electromagnetic spectrum is visible. what determines which bands of the electromagnetic spec
swat32

Answer:

hsnzbssj

Explanation:

<h2>skibbbbbbbbbbbbibi</h2>
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