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anzhelika [568]

3 years ago

5

An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g

ravitational potential energy of the object, in kJ, relative to the surface of the earth.

1 answer:

Harlamova29_29 [7]3 years ago

4 0

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

$Gravitational Potential Energy = mass\times g\times Height$

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = $9.78m/s^{2}$

Applying values in the equation we get

$Gravitational Potential Energy=251X9.78X24 Joules$

$Gravitational Potential Energy=58914.72 Joules$

$Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ$

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The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

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Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014$

0.14% probability of observing more than 4 errors in the carpet

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3 years ago

If you log into the admin account on windows 10, will the admin be notified ?

professor190 [17]

Answer:

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How may a Professional Engineer provide notice of licensure to clients?

azamat

Find full question attached

Answer:

(b) By including a statement that he or she is licensed by the Board for Professional Engineers and Land Surveyors immediately above the signature line in at least 12 point type on all contracts for services

Explanation:

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The engineer might provide statement to each client to show he is licensed which would then be signed by the client

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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

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$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

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D = 6 cm = 0.06 m

$V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s$

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$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

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Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

For the pipe 2, the flow velocity is:

$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087$

The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

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$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

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$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

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