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3 years ago

Which layer of the sun is responsible for producing the light shown in the picture above?

Physics

2 answers:

jasenka [17]3 years ago

7 0

It is photosphere, so your answer would be C. I took the quiz and this was the answer.

The photosphere basically radiates light into space, that's why that's the answer! Have a great day!

Eva8 [605]3 years ago

3 0

Answer:

C. Photosphere

Explanation:

The lights shown in the figure comes from the outermost layer of the Sun. This layer is called photosphere.

This is the layer from where the light of the Sun is radiated, before travelling through space and reaching us.

The photosphere is the coldest layer of the Sun: its surface temperature is between 4500 and 6000 K. Its width is approximately 100 km.

A characteristic of the photosphere is the presence of the sunspots, which appear as darker spots, and are regions of lower temperature caused by a concentration of magnetic flux.

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A series RL circuit includes a 6.05 V 6.05 V battery, a resistance of R = 0.655 Ω , R=0.655 Ω, and an inductance of L = 2.55 H.

Ivenika [448]

Answer:

The induced emf 1.43 s after the circuit is closed is 4.19 V

Explanation:

The current equation in LR circuit is :

$I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })$ .....(1)

Here I is current, V is source voltage, R is resistance, L is inductance and t is time.

The induced emf is determine by the equation :

$V_{e}=L\frac{dI}{dt}$

Differentiating equation (1) with respect to time and put in above equation.

$V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }$

$V_{e}=Ve^{\frac{-Rt}{L} }$

Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.

$V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }$

$V_{e}=4.19\ V$

5 0

3 years ago

A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu

loris [4]

According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10⁻¹² C
r = 5.0 cm = 0.05 m

Now, we can apply the formula:
E(r) = k · (q₁ + q₂) / r²
= 9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C

Hence, the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C

4 0

4 years ago

A spring with a constant of 92N/m is compressed 2.8 cm. How much potential energy is stored in the spring?

Arada [10]

Answer:

0.036J

Explanation:

Given parameters:

Spring constant , K = 92N/m

Compression = 2.8cm = 0.028m

Unknown:

Potential energy = ?