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3 years ago

I’LL MARK YOU BRAINIEST IF U ANSWER IT

Physics

2 answers:

Kryger [21]3 years ago

6 0

4 is the answer to this question

Sidana [21]3 years ago

4 0

Answer:

2is the answer to this question

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A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

erastova [34]

Gpe = mgh
gpe = 50*10*3
gpe = 1500 J

7 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

THIS IS MY EXAM HURRY PLS

tangare [24]

Answer:

elements in the same column have the same number of neutrons. elements with similar mass are placed in the same column.

5 0

3 years ago

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Which of the following statements it's true about Sir isaac Newton

krek1111 [17]

The legend is that he discovered gravity when an apple feel on his head. I don’t know what the true story is, but that’s what I’ve heard so maybe A??
Although, I’m pretty sure it could also be C
So... between A and C, however, I don’t want you to get it wrong so I would recommend getting another opinion
Hope this helps!

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3 years ago

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The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

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3 years ago

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