WHICH COMPONENT IN A BLOWER MOTOR

A teacher tells her students, "When you do your math homework assignments, you must use white lined paper. Please, no tear-o

Sindrei [870]

Answer:

The white lined paper

Explanation:

The teacher is most likely putting the while line paper in jeopardy because of the detail process involved in taking care of the paper prior to the submission of the home work.

The fact that a mistake must not be visible due to the instruction of every erasures being thorough and clean. this can cause jeopardy to the paper.

8 0

3 years ago

What is the focus of 7th grade civics

Jobisdone [24]

C. seems like the best answer. i may be wrong so don’t quote me on that

7 0

3 years ago

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Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload

Vinil7 [7]

Answer:

1) $V_o = 10 liters$

2) $V_o = 12.26 liters$

Explanation:

For isothermal process n =1

$V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}$

$V_o = 10 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.03

$actual \ volume = c1\times 10 = 10.3 liters$

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

$V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}$

$V_o = 12.26 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.15

$actual \volume = c1\times 10 = 11.5 liters$

8 0

3 years ago

I accidently peed my pants help me change me pls

nataly862011 [7]

Answer: *changed*

Explanation: Because you peed

3 0

3 years ago

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From an aerial photograph, one observes that on a level section of a (multilane) highway, 25% of the vehicles are trucks, 75% ar

egoroff_w [7]

Answer:

(a) Flow rate of vehicles = No of vehicles per mile * Speed

=No of cars per mile * Speed +No of trucks per mile * Speed

= 0.75*50*60 + 0.25*50*40

=2750 vehicles / hour

(b) Let Density of vehicles on grade = x

Density on flat * Speed =Density on grade * Speed

So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25

So, x= 57.89

So, Density is around 58 Vehicles per Mile.

(c) Percentage of truck by aerial photo = 25%

(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %

4 0

3 years ago