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Kazeer [188]
3 years ago
5

What would be the diameter of the earth if it had the density of a neurton star

Physics
1 answer:
Marat540 [252]3 years ago
5 0
The entire mass of the Earth at neutron star density would fit into a sphere of 305 m in diameter
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Someone plzzz help it’s my test it’s 50 summative points I’ll mean a lot of you can help thank youuu!!
DaniilM [7]

Answer:if you go to g○○gle there's a camera icon next to the search bar, click on that and take a photo of your test, then they'll find the answers that other people answered or needed, hope that helps

Explanation:

8 0
3 years ago
Read 2 more answers
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m^2 . The wave is incident
Alchen [17]

Answer:

47 mW

Explanation:

The average value of the Poynting vector, S = 0.939 W/m² = Intensity of wave, I

S = I S

Also, I = P/A where P = Et, P = power of electromagnetic wave, E = energy of electromagnetic wave in time t and t = time = 1 min = 60 s and A = area = lb since the electromagnetic waves falls on area equal to that of a rectangle.

So, S = Et/A

E = SA/t

= Slb/t

= 0.939 W/m² × 1.5 m × 2.0 m/60 s

= 2.817 W/60 s

= 0.047 W

= 47 mW

So, 47 mW of electromagnetic energy falls on the area in 1.0 minute.

4 0
3 years ago
Law of motion that says for every action there is an equal and opposite
hodyreva [135]

Answer: Newton's third law

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

Explanation:

4 0
3 years ago
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You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of
Rzqust [24]

Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

                                       = 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0
3 years ago
A force p is 46.67N. if it inclined at an angle of 50°, find its horizontal component​
Ivanshal [37]

Explanation:

F_{x} = 46.67 N(cos 50°) = 30.0 N

8 0
3 years ago
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