The moment of intertia about the z-axis of the solid shown on the right with density deltaδequals=1 is upper i subscript z base
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Answer:1.69*10^12 J
Step-by-step explanation:
From figure above, using triangle ratio
485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485
Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3
Force slice = 150*755.5^2.y^2.∆y/485^2
From figure 2 in the attachment work done for elementary sclice
Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2
= (150*755.5^2*y^2)(485-y)∆y/485
To calculate the total work we integrate from y=0 to y= 485
Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485
Integrating the above
W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485
W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)
Work done 1.69*10^12joules
Answer:
The degrees of freedom are given by:
The p value for this case would be given by:
Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes
Step-by-step explanation:
Information given
represent the sample size
represent the confidence level
represent the sample variance
represent the value that we want to verify
System of hypothesis
We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:
Null Hypothesis:
Alternative hypothesis:
The statistic is given by:
And replacing we got:
The degrees of freedom are given by:
The p value for this case would be given by:
Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes