We need to define the variables,
So,
![F_x (x) = 1-e^{-\lambda x}\\F_x (x) = 1-e^{-0.5x}](https://tex.z-dn.net/?f=F_x%20%28x%29%20%3D%201-e%5E%7B-%5Clambda%20x%7D%5C%5CF_x%20%28x%29%20%3D%201-e%5E%7B-0.5x%7D)
Therefore, the probability that the repair time is more than 4 horus can be calculate as,
![P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018](https://tex.z-dn.net/?f=P%28x%3E4%29%3D1-P%28x%3C4%29%5C%5CP%28x%3E4%29%3D%201-F_x%284%29%5C%5CP%28x%3E4%29%20%3D%201-e%5E%7B-0.5%2A4%7D%5C%5CP%28x%3E4%29%20%3D%201-0.98%5C%5CP%28x%3E4%29%20%3D%200.018)
The probability that the repair time is more than 4 hours is 0.136
b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,
![P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}](https://tex.z-dn.net/?f=P%28x%5Cgeq%2012%7Cx%3E7%29%3DP%28X%5Cgeq7%2B5%7Cx%3E7%29%5C%5CP%28x%5Cgeq12%7Cx%3E7%29%3DP%28X%5Cgeq5%29%5C%5CP%28x%5Cgeq12%7Cx%3E7%29%3D1-P%28x%5Cleq%205%29%5C%5CP%28x%5Cgeq12%7Cx%3E7%29%3D1-e%5E%7B-0.5%282%29%7D)
![P(x\geq 12|x>7)=0.6321](https://tex.z-dn.net/?f=P%28x%5Cgeq%2012%7Cx%3E7%29%3D0.6321)
The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63
Answer: smallest will be 000,.111 84.
Explanation:
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1
a) 0.01 μF
b) 0.58 μF
c) 0.060 μF
d) 0.8 μF
Answer:
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
Explanation:
Please refer to the attached Figure 12-1 where three capacitors are connected in series.
We are asked to find out the equivalent capacitance of this circuit.
Recall that the equivalent capacitance in series is given by
![$ \frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} $](https://tex.z-dn.net/?f=%24%20%5Cfrac%7B1%7D%7BC_%7Beq%7D%7D%20%3D%20%20%5Cfrac%7B1%7D%7BC_%7B1%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BC_%7B2%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BC_%7B3%7D%7D%20%24)
Where C₁, C₂, and C₃ are the individual capacitance connected in series.
C₁ = 0.1 μF
C₂ = 0.22 μF
C₃ = 0.47 μF
So the equivalent capacitance is
![$ \frac{1}{C_{eq}} = \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47} $](https://tex.z-dn.net/?f=%24%20%5Cfrac%7B1%7D%7BC_%7Beq%7D%7D%20%3D%20%20%5Cfrac%7B1%7D%7B0.1%7D%20%2B%20%5Cfrac%7B1%7D%7B0.22%7D%20%2B%20%5Cfrac%7B1%7D%7B0.47%7D%20%24)
![$ \frac{1}{C_{eq}} = \frac{8620}{517} $](https://tex.z-dn.net/?f=%24%20%5Cfrac%7B1%7D%7BC_%7Beq%7D%7D%20%3D%20%20%5Cfrac%7B8620%7D%7B517%7D%20%20%24)
![$ C_{eq} = \frac{517}{8620} $](https://tex.z-dn.net/?f=%24%20C_%7Beq%7D%20%3D%20%20%5Cfrac%7B517%7D%7B8620%7D%20%20%24)
![$ C_{eq} = 0.0599 $](https://tex.z-dn.net/?f=%24%20C_%7Beq%7D%20%3D%20%200.0599%20%20%24)
Rounding off yields
![$ C_{eq} = 0.060 \: \mu F $](https://tex.z-dn.net/?f=%24%20C_%7Beq%7D%20%3D%20%200.060%20%5C%3A%20%5Cmu%20F%20%24)
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
Answer:
the restoring force is = 3/4NKT
Explanation:
check the attached files for answer.
Answer:
See explanation
Explanation:
The magnetic force is
F = qvB sin θ
We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields
F
=
(
20
×
10
−
9
C
)
(
10
m/s
)
(
5
×
10
−
5
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=
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×
10
−
11
(
C
⋅
m/s
)
(
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=
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×
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−
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N