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Orlov [11]
3 years ago
10

Dimensional analysis: a)-Can be used to derive equations and non-dimensional numbers. b)-Helps in planning experiments. c)-Is an

efficient way of storing and retrieving experimental data. d)-Is very helpful in determining speeds for dynamic similarity in model testing. e)- All of the above
Engineering
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

Option e)

Explanation:

Dimensional Analysis helps in the analysis of relationship between physical quantities, checking its dimension thus checking the accuracy of a given or to find the dimensions of a quantity or the degree of correctness of an equation.

Dimensional analysis has some of the applications that are listed below:

  • To establish a relationship between physical quantities
  • To check the accuracy of a given formula
  • To derive units of a physical quantity
  • To convert system of units in one another.
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anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

4 0
3 years ago
Select the best answer for the question.
zepelin [54]
The answer for this question is A
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If superheated water vapor at 30 MPa iscooled at ​constant pressure​, it will eventually become saturated vapor, and with suffic
nirvana33 [79]

Answer:

False.

Explanation:

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We have a tube with a diameter of 5 inches that is 1 foot long. The tube then reduces the diameter to 3 inches. According to the
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Answer:

correct option is (A) 0.5

Explanation:

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solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

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N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

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