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arsen [322]
3 years ago
10

Suppose electric power is supplied from two independent sources which work with probabilities 0.4, 0.5, respectively. if both so

urces are providing power enough power will be available with probability 1. if exactly one the them works there will be enough power with probability 0.6. of course, if none of them works the probability that there will be sufficient supply is 0.a) what are the probabilities that exactly k sources work for k=0,1,2?b)compute the probability that enough power will be available.
Physics
1 answer:
sattari [20]3 years ago
3 0

Answer:

The answers to the questions are as follows

a)  k = 0, P = 0.3

k = 1, P = 0.5

k = 0, P = 0.2

b) The probability that enough power will be available is 0.5.

Explanation:

To solve the question we write the parameters as follows

Probability that the first power source works = P(A) = 0.4

Probability that the second power source works = P(B) = 0.5

When both sources are supplying power  we have the probability = 1

If non of them is producing the probability = 0

a) The probability that exactly k sources work for k=0,1,2 is given by

For k = 0, probability = (1- P(A))× (1- P(B)) = 0.6 × 0.5 =0.3

Therefore the probabilities that exactly 0 source work  = 0.3

for k = 1 we have the probability = P(A)(1-P(B)) + P(B)(1-P(A)

= 0.4(1-0.5)+0.5(1-0.4) = 0.2 + 0.3 = 0.5

The probabilities that exactly 1 source work  = 0.5

for k = 2 we have the probability given by = P(A) × P(B) = 0.4 × 0.5 = 0.2

Therefore the  probability that exactly 2 sources work  = 0.2

b)  The probability that enough power will be available is

0 × P(k = 0) + 0.6 × P(k = 1) + 1 × P(k = 2)

0 × 0.2 + 0.6 × 0.5 + 1 × 0.2 = 0.5

The probability that enough power will be available is 0.5.

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If something weighs 2891N on earth, what’s it’s mass?
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Answer:

289.1kg

Explanation:

because W=M*G, and M=W/G, so 2891N/10m/s²=289.1kg

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(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 12
Elodia [21]

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

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Which poison was used in the Jonestown massacre?
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Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardr
Margaret [11]

Answer:

a. d₁/d₂ = 1.09 b. 0.054 mW

Explanation:

a. What is the ratio of the diameter of the first student's eardrum to that of the second student?

We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So, I ∝ I/d²

I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.

Given that I₂ = 1.18I₁

I₂/I₁ = 1.18

Since I₁/I₂ = d₂²/d₁²

√(I₁/I₂) = d₂/d₁

d₁/d₂ = √(I₂/I₁)

d₁/d₂ = √1.18

d₁/d₂ = 1.09

So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09

b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

We know intensity, I = P/A where P = acoustic power and A = area = πd²/4

Now, P = IA

= I₂A₂

= I₂πd₂²/4

= 1.18I₁πd₂²/4

Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

So, P = 1.18I₁πd₂²/4

= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

= 0.691244π × 10⁻⁴ W/4 =

2.172 × 10⁻⁴ W/4

= 0.543 × 10⁻⁴ W

= 0.0543 × 10⁻³ W

= 0.0543 mW

≅ 0.054 mW

3 0
3 years ago
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