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11111nata11111 [884]
3 years ago
11

In comparing sunspots to other areas of the sun's surface one can say that sunspots are...?

Physics
1 answer:
taurus [48]3 years ago
8 0

Answer:

D. Cooler

Explanation:

The Sun's surface has a temperature ranging from 6 000 Kelvin, thus it is very hot. The radioactive reaction in the core of the Sun is a nuclear fusion reaction. This reaction ensures continuous release of high energy from the surface of the Sun.

But during the reaction, some parts becomes cooler than other parts on its surface. Which is due to the release of high amount of energy into space. The Sun's spot can be found in the cooler part of the Sun.

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Charge is moved of a capacitor

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When Arti kicks a football, two forces interact. Arti's foot exerts a force on the ball. What exerts a force on Arti's foot?
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The football and air resistance between contact
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What is one joule work
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What would be the answer for this and how?
Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

k = \frac{F}{\Delta x}\\

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\

For a load of 30 N, applying Hooke's Law:

\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\

Hence, the correct option is:

<u>B. 6 cm</u>

7 0
3 years ago
A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil
Travka [436]

Complete question:

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A.

What is magnitude of the magnetic moment of the coil? Answer in units of A · m2.

Answer:

The magnetic moment of the coil is 0.0252 A.m²

Explanation:

Given;

radius of the coil, r = 4 cm = 0.04 m

number of turns of the coil, N = 5 turns

magnetic field strength B = 0.8 T

current in the coil, I = 1 A

Area of the coil, A = πr² = π(0.04)² = 0.00503 m²

magnetic moment of the coil, μ = NIA

where;

N is the number of turns

I is the current in the coil

A is the area of the coil

magnetic moment of the coil, μ = 5 x 1 x 0.00503 = 0.0252 A.m²

Therefore, the magnetic moment of the coil is 0.0252 A.m²

8 0
3 years ago
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