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Umnica [9.8K]
2 years ago
15

Explain how a charged amber rod can lift tiny bits of paper despite being under the gravitational force of the entire mass of th

e earth.
Physics
1 answer:
nika2105 [10]2 years ago
4 0

Explanation:

When a charged amber rod is brought near the tiny bits of paper. The bits of paper are neutral. There is an electrostatic force of attraction between the amber rod and tiny bits of paper.

The charged amber rod attracts or repel the bits of paper depending on the charge of amber rod. The electrostatic force is much stronger than the gravitational force.

Therefore, a charged amber rod can lift tiny bits of paper despite being under the gravitational force of the entire mass of the earth.

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Friction between the ball and the floor is stealing some of the kinetic energy of the ball, and turning it into heat.

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Which option correctly matches the chemical formula of a compound with its name?
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2 years ago
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which one of these represent total momentum of a system of two particles traveling one against the other?
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The total momentum of a system is the vector sum of all the individual masses that comprise the system.

Moreover, To calculate the total momentum of two objects during a collision, add their individual momentums. You can calculate the momentum for each object using the formula p=mv, where p is the momentum, m is the mass, and v is the velocity. The law of conservation of momentum can be expressed as follows. For a collision between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

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5 0
11 months ago
The Graph above shows the speed of a car traveling in a straight line as a function of time. The car accelerates uniformly and r
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Answer:

Where is the graph??

If a car travels from zero to 4 m/s ins 8 sec

a = 4 / 8 = .5 m/s^2

V (2) = 2 * .5 = 1 m/s after 2 sec

S = V t + 1/2 a t^2

S = 1 * 3.9 + 1/2 * 1/2 *3.9^2 = 3.9 + 3.80 = 7.70 m   from 2 to 5.9  sec

Check:

Total distance traveled = 1/2 a t^2 = 5.9^2 / 4 = 8.70 m

Distance traveled in 2 sec = 1/2 * 1/2 * 4 = 1 m

Total distance from 2 to 5.9 = 8.7 - 1 = 7.7 m agreeing with thw above

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