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IgorC [24]
3 years ago
12

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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3 years ago
You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
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Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

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2 years ago
Technician A states that a brake lathe is used to make a used brake rotor surface "like new". Technician B states that a brake l
nikitadnepr [17]

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Both Technician A and B are correct.

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2 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
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The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

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