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Morgarella [4.7K]
2 years ago
8

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

ibility that (a) all 4 are defective? (b) 3 are defective and 1 is good? (c) exactly 2 are defective? (d) none are defective?
Physics
1 answer:
dlinn [17]2 years ago
8 0

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) C(4,4) = 1; C(15,4) = 1365

P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326

b) C(4,3) = 4; C(11,1) = 11

P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223

c) C(4,2) = 6; C(11,2) = 55

P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418

d) C(11,4) = 330

P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418

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Answer:

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When an additional resistance of 3.4 Ω is inserted in series with R,

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Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I'  and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

1 = 1.335R/(R+3.4)

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Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

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(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

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If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

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(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

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