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AlladinOne [14]
3 years ago
13

A team of students conducts a series of experiments to investigate collisions. In the first experiment, the two carts collide wi

th each other on a smooth surface. The carts stick together and continue to move forward. In the second experiment, the two carts collide with each other on a rough surface. The carts stick together and quickly come to rest. In both experiments, the initial speeds of the carts are identical. Is there a difference in the total energy of the two experiments?
Physics
1 answer:
never [62]3 years ago
7 0

Answer: Yes, there is a difference in the total energy of the two experiments because the first two carts that were on a smooth surface had friction which causes two or more objects to stick together.

Explanation:

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A +0.2 µC charge is in an electric field. What happens if that charge is replaced by a +0.4 µC charge?
AleksandrR [38]

Explanation:

It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.

So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.

                 U = qV

Since, the electric potential is directly proportional to the charge. Hence, when 0.2 \mu C tends to replaced by 0.4 \mu C then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.

Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.

4 0
3 years ago
Explain why the types of technology valued can vary.
Elan Coil [88]
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4 0
3 years ago
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Which of the following statements are true about the international system of measurement?
anygoal [31]
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8 0
3 years ago
An object is located 70 cm from a concave mirror with a focal length of 15 cm. What is the image
Stels [109]

(a) The distance of the image formed by the concave mirror is 19.1 cm.

(b) The image formed is diminished and real.

<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/15 - 1/70

1/v = 0.05238

v = 1/0.05238

v = 19.1 cm

The image distance is smaller than object distance, thus the image formed is diminished and real.

Learn more about concave mirror here: brainly.com/question/13164847

#SPJ1

5 0
1 year ago
I NEED THIS ASAP PLEASE
kolbaska11 [484]

Answer:

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Explanation:

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