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3 years ago

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A diagonal aluminum alloy tension rod of diameter d and initial length l is used in a rectangular frame to prevent collapse. The

rod can safely support a tensile stress of sallow. If d 5 0.5 in, l 5 8 ft, and sallow 5 20 kpsi, determine how much the rod must be stretched to develop this allowable stress.

1 answer:

Ksenya-84 [330]3 years ago

6 0

Answer:

$\Delta L = 0.1883 inch$

Explanation:

Given data:

d = 0.5 inch

$L_i = 8 ft$

$\sigma_{allow} = 20 kpsi$

we know that change in length is given as

$\Delta L = \frac{PL}{AE}$

$= \frac{P}{A}\times \frac{L}{E}$

$= \sigma_{allow} \times \frac{L}{E}$

modulus of elasticity E for aluminium alloy is $10.2 \times 10^6 psi = 10.2 \times 10^3 kpsi$

$\Delta L = \frac{20 \times 8}{10.2\times 10^3}$

$\Delta L = 0.01569 ft$

$\Delta L = 0.1883 inch$

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A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

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Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

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3 years ago

When would working with machinery be a common type of caught-in and caught-between<br> hazard?

tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.

Explanation:

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3 years ago

Multiple Choice

Ymorist [56]

Answer:

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Generally, when we have water leaks in almost any building or equipment, we use a sealant. However, this sealant could be of different types depending on the peculiarity of the leakage.

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son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

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we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

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Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

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Answer:

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