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Stolb23 [73]
2 years ago
13

Eventually, the process approaches a steady state. In that steady state, the charge of the capacitor is not changing. What is th

e current in the circuit in the steady state?
Physics
1 answer:
devlian [24]2 years ago
6 0

Answer:

The current in the circuit must be zero.

Explanation:

In a RC circuit, the steady state is reached when either the capacitor is fully charged or fully discharged. In either case, there must not be any current through the circuit because if it exists, it will deliver charge to the capacitor and thus change its charge, which is not a steady state.

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I need help please somebody help me
Morgarella [4.7K]

Answer:

Generally, magnets are attracted to objects that are made of the metals iron, nickel, or cobalt. These materials are called ferromagnetic materials. ... When all or most of the domains are aligned in the same direction, the whole object becomes magnetized in that direction and becomes a magnet.

Explanation:

5 0
2 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

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