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pantera1 [17]
3 years ago
6

A light-year is:

Physics
1 answer:
Elina [12.6K]3 years ago
4 0
A is the answer for this question
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Liquids with convex meniscus​
kobusy [5.1K]

Mercury in a glass tube.

7 0
2 years ago
What is the speed of each neutron as they crash together? keep in mind that both neutrons are moving?
Sophie [7]
This involves shooting electrons (from an accelerator) at a target or protons. This technique provided evidence for the existence of quarks. <span>proton-antiproton scattering as well.
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7 0
3 years ago
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If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a resul
Kobotan [32]

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy.  So the <em>kinetic energy</em> in the system <em>must</em> <em>decrease</em>.

In fact, this isn't even a "result".  The kinetic energy has to decrease <em><u>before</u></em> the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out.  If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

7 0
3 years ago
Can u Anwser this Plzzzz
nata0808 [166]

Answer:

-0. 75m/s^2

Explanation:

use formula of acceleration

5 0
3 years ago
Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2
Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

F_n=F-F'

F_n=37.2-22.4

F_n=14.8\ N

Work done on the wagon is given by the product of net force and displacement. It is given by :

W=F_n\times d

W=14.8\ N\times 2.5\ m

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
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