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3 years ago

6

What is the differences between total revenue and total costs? Make

1 answer:

solong [7]3 years ago

6 0

Answer:

The basic difference between Total cost and total revenue is that the total cost includes the total expenditure incurred on the production of a commodity whereas total revenue refers to the money received from selling that commodity.

Explanation:

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forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fue

Zina [86]

Answer: 131.75minutes

Explanation:

First if all, we've to find the density of liquid which will be:

= Specific gravity × Density to pure water

= 0.91 × 8.34lb/gallon

= 7.59lb/gallon

Then, the volume that's required to fill the tank will be:

= Load limit/Density of fluid

= 40000/7.59

= 5270.1gallon

Now, the time taken will be:

= V/F

= 5270.1/40

= 131.75min

It'll take 131.75 minutes to fill the tank in the truck.

5 0

3 years ago

Which of the following is lost when the computer is turned off? ILL MARK BRAINLEST

Pavel [41]

Answer:

cloud storage is the answer

6 0

3 years ago

Read 2 more answers

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

a)

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

b)

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago

Tech A says you can find the typical angle of a V-block engine by dividing the number of cylinders by 720

jeka57 [31]

Answer:

Tech A is correct

Explanation:

Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.

Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.

6 0

3 years ago

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

3 years ago