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Olin [163]
3 years ago
13

How does the electrostatic force compare with the strong nuclear force in the

Physics
2 answers:
soldier1979 [14.2K]3 years ago
7 0

Answer:

The electrostatic force is weaker and acts over longe distances, is the correct answer for Appex/

Explanation:

diamong [38]3 years ago
3 0

Answer:

Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force

Explanation:

There are mainly two forces acting between protons and neutrons in the nucleus:

- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by

F_E=\frac{kq_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 are the  charges of the two particles, r is the separation between the particles.

The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.

- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.

At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.

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When a falling object reaches terminal velocity, the net force acting on it is
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Kelly is riding a bicycle, moves with an initial velocity of 5 m/s. Ten seconds later, she is moving at 15 m/s. What is her acce
sladkih [1.3K]

Answer:

her acceleration is 1 m/sec

Explanation:

The following information is given in the question

The initial velocity is 5 m/s

After 10 seconds, she would be moved at 15 m/s

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The same would be considered  

3 0
3 years ago
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
Artemon [7]

Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

4 0
3 years ago
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