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nirvana33 [79]
3 years ago
13

A 20.0 μF capacitor initially charged to 30.0 μC is discharged through a 1.20 kΩ resistor. How long does it take to reduce the c

apacitor's charge to 15.0 μC
Physics
1 answer:
Kay [80]3 years ago
3 0

Answer:

16.6 ms or 0.0166 s

Explanation:

If Q is the final charge, Q' is the initial charge, C the capacitance ,R is the resistance , t the time taken  and τ the time constant,

[tex]Q = Q'( 1- e^{-t\div \tau })

τ = R C = (1.20×10³) (20×10⁻⁶) = 0.024 s

15 = 30 ( 1- e^{-t\div \ 0.024 })

( 1- e^{-t\div \ 0.024 }) = 15 ÷ 30

⇒ - e^{-t\div \ 0.024 }) = 0.5 -1  

⇒  e^{-t\div \ 0.024 }) = 0.5

Taking logarithm to the base e on both sides of this equation,

⇒ t = 0.0166 seconds = 16.6 milli seconds

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\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}

\\ \rm\Rrightarrow u=\dfrac{190}{-14}

\\ \rm\Rrightarrow u=13.6cm

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5 0
2 years ago
** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th
Vlad [161]

As we know that in transformers we have

\frac{V_s}{V_p} = \frac{N_s}{N_p}

here we know that

V_s = 17,500 Volts

V_p = 350,000 Volts

N_s = 600 coils

now from above equation we will have

\frac{17500}{350000} = \frac{600}{N_p}

N_p = 600\times \frac{350000}{17500}

N_p = 12000 coils

6 0
3 years ago
Would it be better to collide head-on with an identical car traveling at the same speed or collide with a stationary brick wall?
attashe74 [19]
It makes no difference. The momentum of either car goes to zero in both cases.
6 0
3 years ago
Doe anyone get this ​
8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

  • 2 m/s²
  • 0.19 m/s²
  • 9.25 m/ s²
  • 0.04 m/s²
  • 100.39 m/s²

4 0
3 years ago
An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is
Ulleksa [173]

Answer:

1. a W_t=746.63 J

  b E_p=212.415 J

  c W_n=183.96J

2. T_e=99.71J

Explanation:

a). The work done by the tension is:

W_t=T*dt

dt=\frac{5.1}{cos(17)}

W_t=140N*\frac{5.1}{cos(17)}

W_t=746.63 J

b). The work done potential of gravity

E_p=m*g*h

h=5.1*sin(17)

E_p=8.5kh*9.8*5.1*sin(30)

E_p=212.415 J

c). The work done by the normal force

W_n=N*d_n

d_n=5.1*sin(30)=2.55

W_n=8.5kg*9.8*cos(30)*2.55

W_n=183.96J

2. The increase in thermal energy is:

T_e=F*d

F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)

F_k=19.5N

T_e=19.55*5.1m

T_e=99.71J

4 0
3 years ago
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