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3 years ago

Show that -40 F is approximately equal to -40 C.

Engineering

1 answer:

V125BC [204]3 years ago

8 0

Answer: Please see below the explanation.

Explanation:

The Celsius (ºC) and Farenheit (ºF), are different temperature scales, but both have identifiable values for two important physical phenomena, water freezing point and water boiling point.

For º C, these are follows:

Freezing Point= 0º C Boiling Point: 100ºC

For º F, we have the following:

Freezing Point= 32º F Boiling Point: 212 ºF

Assuming that all steps (in each scale) between two following values are equal, we have 100 º C between both points in ºC, and 190º in ºF, so degrees in Farenheit are "shorter" than in Celsius, being the relationship, approximately the following:

1 º C = 9/5 ºF.

Taking into account that, by definition, 0ºC = 32º F, we can use the following equations in order to convert ºC to ºF, and viceversa:

ºC = (ºF-32) 5/9

ºF = 32 + 9/5ºC

Now, let's calculate -40ºF, in ºC:

ºC= (-40-32)* 5/9 = -40º C

Let's do the same, converting -40ºC to ºF:

ºF = 32 + (-40)*9/5 = 32 + (-72) = -40ºF.

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Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

$V(t) = 5(1-e^{-0.5t})$

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$i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)$

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

$i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\$

$i(t) = 1000t +2000e^{-0.5t} -2000\\$

So the current at t = 2 seconds is

$i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A$

The energy stored in the inductor at t = 2 seconds is

$W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J$

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Answer:

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Explanation:

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The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

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$P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}$

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

$\eta_t$ = The efficiency of the pump = 78%

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