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3 years ago

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Se requiere un permiso aprobación o restricción contaminante para todos los métodos comerciales de descarga de aguas residuales

?

1 answer:

OLga [1]3 years ago

4 0

Answer:

sí, necesita una licencia para distribuir productos químicos

Explanation:

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A small pad subjected to a shearing force is deformed at the top of the pad 0.12 in. The heigfit of the pad is 1.15 in. What is

ser-zykov [4K]

Answer:

Shearing strain will be 0.1039 radian

Explanation:

We have given change in length $\Delta L=0.12inch$

Length of the pad L = 1.15 inch

We have to find the shearing strain

Shearing strain is given by

$\alpha =tan^{-1}\frac{\Delta L}{L}=tan^{-1}\frac{0.12}{1.15}=5.9571^{\circ}$

Shearing strain is always in radian so we have to change angle in radian

So $5.9571\times \frac{\pi }{180}=0.1039radian$

6 0

3 years ago

Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are

borishaifa [10]

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

4 0

3 years ago

When checking the resistance of a dual voltage wye motor, there should be ____ resistance readings. 1) twelve 2) six 3) three

hjlf

Answer:

1) twelve

Explanation:

The dual voltage motors are used in day to day operations. The wye is connected with 9 lead motors. Maximum resistance can be obtained if the resistance are connected in series. To check resistance of dual voltage wye motor there must be twelve resistance readings of 1 ohm each.

8 0

3 years ago

If engineering is easy then why don't most people join?

Mazyrski [523]

Engineering requires a lot of school

5 0

3 years ago

Read 2 more answers

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

Oxana [17]

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.

8 0

3 years ago