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just olya [345]
4 years ago
10

. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the

voltage across the capacitor 0.16 seconds after the circuit is connected to the battery
Physics
1 answer:
Licemer1 [7]4 years ago
3 0

Answer:

The  voltage is V_c  = 9.92 \ V

Explanation:

From the question we are told that

     The voltage of the battery is  V_b  =  24 \ V

     The capacitance of the capacitor is  C  =  3.0 mF  =  3.0 *10^{-3} \  F

     The  resistance of the resistor is R   =  100\  \Omega

     The time taken is  t =  0.16 \ s  

Generally the voltage of a charging charging capacitor after time t is mathematically represented as

       V_c  =  V_o (1 -  e^{- \frac{t}{RC} })

Here V_o is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so  

      V_c  =  24 (1 -  e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })

      V_c  = 9.92 \ V

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5 0
3 years ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
4 years ago
if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *
SSSSS [86.1K]

Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

F=0.013 N

4 0
3 years ago
A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13
tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

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here we have

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here we know that

F = 4.8 \times 10^{-13} N

q = 1.6 \times 10^{-19} C

v = 4 \times 10^6 m/s

now from above equation we have

B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}

B = 0.75 T

8 0
4 years ago
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