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Vaselesa [24]
3 years ago
6

A 2.34 kg book is dropped from a height of +3.1 m. (a) What is its acceleration?

Physics
1 answer:
frosja888 [35]3 years ago
8 0
<span> 9.8 m/s^2 is the answer</span>
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A box is pulled with a horizontal force of 500N and moves 5m what is work done
dalvyx [7]
The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

4 0
3 years ago
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

  • Q, the charge stored in the capacitor, and
  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

  • \epsilon is the permittivity of the material between the two plates.
  • A is the area of each of the two plates.
  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

V = \displaystyle \frac{Q}{C}.

The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

3 0
3 years ago
NEED HELP ASAP!!!
sineoko [7]

Answer:

2,500 watts

Explanation:

I got this answer right on a test. I hope it works for you to.

5 0
3 years ago
Select the options that best complete the statement.
andriy [413]

Answer:

Positively charged particle trajectories always follow electric field lines because the electric force on a positively charged particle is in the same direction as the electric field.

Explanation:

For any positive charge the electric field emerges radially outwards and it goes radially inwards for the negative charges.

  • From the theory of electric field lines we know that they never intersect each other, either they get merged when the sources are unlike or they repel when the sources are alike. In other words the electric field lines align in the same direction as that of the field.
  • So, when a positive charge is released into the an electric field they follow the direction of the field lines because they too have their field lines emerging radially outwards and hence these lines align in the direction of the field.

8 0
3 years ago
To remove 700 j of heat from a refrigerator, the compressor in the refrigerator does 500 j of work. How much heat is released in
Vinvika [58]
If the compressor removes 500 j if the 700 the 200 j left would have Ben conducted through the refrigerator/200 j released into the room
7 0
3 years ago
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