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arsen [322]
2 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]2 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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3 years ago
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

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2 years ago
a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

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Answer:

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Disclaimer: did math in my head, so you better double check with a calculator.

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