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3 years ago

Compare between chromosphere, corona and photosphere

Physics

1 answer:

gladu [14]3 years ago

5 0

Answer:

Explained

Explanation:

Photosphere: The lowest layer of the sun is called photo sphere . It is about 300 miles thick from the surface. It is the source of solar flares. It is marked by bright bubbling granules of plasma.

chromosphere emits a reddish glow as the super heated hydrogen burns off but the red rim can only be seen during total solar eclipse.

The third layer of the sun atmosphere is Corona. It can also only be seen during during a total solar eclipse. Temperature in corona can reach as high as 3.5 million degree fahrenheit. As the gases cool they become solar winds.

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To calculate the heat needed to melt a block of ice at its melting point what do you need to know

Alisiya [41]

You have to know what the temperature of your suroundings first and then you can melt it with the tools you need. yeah that's another thing know what tools you need and then be safe!:D

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3 years ago

A spring attached to a mass is at rest in the initial

svet-max [94.6K]

Answer:

E=(1/2)kx^2

Explanation:

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3 years ago

Read 2 more answers

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

2 years ago

A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa

kotykmax [81]

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is $v_i=180~m/s$ , the final velocity is $v_i=0~m/s$ , and the total time of the motion is $\Delta t=0.02~s$ , so the acceleration is given by
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
$S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m$
So, the bullet penetrates the sandbag 1.8 meters.

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3 years ago

Which of the following is an example of deposition?

Murrr4er [49]

B would be the correct answer

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3 years ago