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Lina20 [59]
3 years ago
9

Select the correct answer. What happens to the sum of the ball’s kinetic energy and potential energy as the ball rolls from poin

t A to point E? Assume there’s no friction between the ball and the ground. A. The sum decreases. B. The sum increases. C. The sum remains the same. D. The sum always equals zero.
Physics
2 answers:
barxatty [35]3 years ago
7 0

Answer:

It might be the sum is the same

Explanation:

Correct me if i am wrong but there is no friction so that means there would not be anything slowing down the ball, therefore the ball wouldn't change or the energy wouldn't change because the ball would never stop.

AlekseyPX3 years ago
3 0

The sum of the ball’s kinetic energy and potential energy remains the same as the ball rolls from point A to point E ( if there’s no friction between the ball and the ground).

Answer: Option C

<u>Explanation: </u>

Based on the law of mechanical energy's conservation, the sum of both kinetic and potential energies i.e. the total amount of mechanical energy remain conserved even in the absence of dissipative forces (friction or air resistance) in a bound system.

                    E = Kinetic + Potential

The kinetic energy is high when ball goes downward but potential energy decreases and reverse happens when ball goes up. But in these case, the sum energy would be constant one.

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nasty-shy [4]

Answer:

Option B. 5 nC

Explanation:

From the question given above, the following data were obtained:

Capicitance (C) = 100 pF

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Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Next, we shall determine the quantity of charge. This can be obtained as follow:

Capicitance (C) = 1×10¯¹⁰ F

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Quantity of charge (Q) =?

Q = CV

Q = 1×10¯¹⁰ × 50

Q = 5×10¯⁹ C

Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:

1 C = 1×10⁹ nC

Therefore,

5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C

5×10¯⁹ C = 5 nC

Thus, the quantity of charge is 5 nC

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3 years ago
Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given tha
pshichka [43]
For resistance we have R=ρ l/a
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conductivity,σ =1/ρ

σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
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