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2 years ago

HELP QUICK WILL GIVE BRAINLEST!!!!

Physics

2 answers:

kondor19780726 [428]2 years ago

5 0

Hi, the answer is C. )A brief eruption of intense high-energy radiation from the sun's surface

This i because as you can see in the image, a radiation is coming from the sun, which is the intence high energy radiation.

Hope this helps.

Romashka-Z-Leto [24]2 years ago

3 0

Hello!

The statement that <span>describes the solar feature shooting off into space labeled C is An envelope of plasma surrounding the Sun.

This is called the Solar Corona. It is the outmost layer of the Sun's atmosphere. The Corona is hotter than the visible surface of the Sun. (1 000 000 - 3 000 000 K). It is composed of Plasma (The fourth state of the matter, similar to the gaseous state, but with electrically charged particles). All structural details of the Solar Corona are derived from the Magnetic Field of the Sun.

Have a nice day!</span>

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A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?

belka [17]

Answer:

117.72kW

Explanation:

Given data

Mass m= 50kg

height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

WD= force * distance

WD= mgx

WD= 50*9.81*2

WD= 981 Joules

Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

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3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

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3 years ago

ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.

vekshin1

Answer:

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2 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

mihalych1998 [28]

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3 years ago

Read 2 more answers

Please give answer with solution

UNO [17]

250 m. for a longer explanation or solution look at this article, i’m sorry.
https://www.quora.com/A-projectile-is-thrown-so-it-travels-a-maximum-range-of-1000m-How-high-will-it-rise

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2 years ago