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goblinko [34]
3 years ago
5

1. How is electric potential energy similar to gravitational potential energy? How is it different? Where will an electron bound

in an atom and have the largest electrical potential energy?
Physics
1 answer:
Korolek [52]3 years ago
7 0
Both ve similar equations 
<span>both are energies of one object w.r.t another </span>
<span>differences- electric pe is due to electrostatic force and gravitational pe is due to gravitational force </span>
<span>electric pe is > than gravitational pe since electrostatic force> gravitational force </span>
<span>electron bound in an atom ll ve largest potential enegy in its ground state. i think hope it helps</span>
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As stream discharge increases, only velocity increases.

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2 years ago
A lamp with a resistance of 60 Ohms is plugged into a 110-Volt household circuit. How much current flows through the lamp?
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The current flowing through the lamp is  1.833A

<h3>What is current?</h3>

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6 0
2 years ago
At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta
aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given  :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.  

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next  

F=|F|cosФp            (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet  

F=|F|cosФp

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<em>As this force is in one direction, we could get its vector as next  </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

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