A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3
1 answer:
The work required to pump the water out of the spout is 5000ft⋅lb
Given
The volume V=12(4ft)(5ft)(6ft)=60ft3,
Then the weight is w = γV = (62.5lb /ft³) (60ft³) = 3750lb.
Since work is force × distance, you need a distance here, and the appropriate one is the distance you must raise the center of mass to get the H2O out of the spout.
Since the center of mass of a uniform triangular area is 13 of the way from any side, the distance is (4ft)/3 so the work is (3750lb) (4ft)/3=5000 ft-lb.
In calculus, the area of a horizontal section yft from the lowest point is (5ft) × (6ft) × y / (4ft) = 7.5y ft2.
We have to lift that slice 4ft−y, so work is
w = (62.5)(4 − y) (7.5y)
dy = (62.5)[15y2−2.5y3]40 = 5000ft⋅lb
The work required to pump the water out of the spout is 5000ft-lb
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We need to conduct a hypothesis in order to determine if the mean is greater than specified value, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
For this case the significance is 1%. So we need to find a critical value in the normal standard distribution who accumulates 0.99 of the area in the left and 0.01 in the right and for this case this critical value is:
Step-by-step explanation:
Notation
represent the sample mean
represent the standard deviation for the population
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean is greater than specified value, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
For this case the significance is 1%. So we need to find a critical value in the normal standard distribution who accumulates 0.99 of the area in the left and 0.01 in the right and for this case this critical value is: