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stealth61 [152]
3 years ago
6

For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average

heat transfer coefficient for the smaller-diameter cylinder is:_______
a. The same as that of the larger-diameter cylinder
b. Larger than that of the larger-diameter cylinder
c. Smaller than that of the larger-diameter cylinder
Engineering
1 answer:
densk [106]3 years ago
3 0
C smaller than that of the larger-diameter cylinder
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What colour is best for radiative heat transfer? a. Black b. Brown c. Blue d. White
GarryVolchara [31]

Answer:

The correct answer is option 'a': Black

Explanation:

As we know that for an object which is black in color it absorbs all the electromagnetic radiation's that are incident on it. Thus if we need to transfer energy to an object by radiation the most suitable color for the process  is black.

In contrast to black color white color is an excellent reflector, reflecting all the incident radiation that may be incident on it hence is the least suitable material for radiative heat transfer.

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Which of the following terms is defined as small bumps and slashes within a fluid power system?
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friction

Explanation:

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7 0
3 years ago
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What other ways could a wildfire be contained, extinguished, or slowed down?
galben [10]
Back burning, starting fires infront of the main fire to prevent the fire from spreading and depleting fuel for the fire, digging trenches so the fire has no where to go, dropping water from planes or helicopters.
3 0
3 years ago
an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required
ASHA 777 [7]

Answer:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

Explanation:

Given;

Resistance, R₁ = 50Ω

Resistance, R₂ = 75Ω

Total resistance, R = (R₁R₂)/(R₁ + R₂)

Total resistance, R = (50 x 75)/(125)

Total resistance, R = 30 Ω

According to ohms law, sum of current in a parallel circuit is given as

I = I₁ + I₂

I = \frac{V}{R_1} + \frac{V}{R_2}

Voltage across each resistor is the same

1.6 = \frac{V}{R_2}  

V = 1.6 x R₂

V = 1.6 x 75

V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

3 0
3 years ago
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Please help ASAP and show all work. Thanks
Shalnov [3]

Answer:

1. --a 10

--b 100 in.

2. --a 4

--b 4

3. --a 2

--b 30 rpm

--c 75 ft.·lb.

4. --a Second class lever

--b 50 lbs.

Explanation:

The Actual Mechanical Advantage, AMA, is given as follows;

AMA = \dfrac{F_R}{F_E}

Where;

F_R = The resistance force magnitude

F_E = The effort force magnitude

1. a. We have;

F_R = 10 lb.

F_E = 100 lb.

AMA = \dfrac{100 \ lb}{10 \ lb} = 10

b. Mechanical \ advantage, \ M.A. = \dfrac{Distance \ moved \ by \ load}{Distance \ moved \ by \ effort}

The diameter of the axle, d = 2 in.

Let 'D' represent the diameter of the wheel, we have;

The distance moved by the axle, c = π·d

The distance moved by the load, C = π·D

M.A. = 10 =  \dfrac{\pi \cdot D}{\pi \times 2}

∴ 2 × 10 = D

D = 20 in.

The required wheel diameter to overcome the resistance force, D = 100 in.

2. --a The mass of the participants, m = 200 lb.

The depth of the ground of the participants = 20 feet

The effort force = 50 lb

Actual Mechanical Advantage, AMA = 200 lb./(50 lb.) = 4

--b. The number of strands of pulley needed ≈ The mechanical advantage = 4

3. The number of gears on Gear A = 10 teeth

The number of gears on Gear B = 8 teeth

The number of gears on Gear C = 20 teeth

-a. Given that the driver gear = Gear A

The output gear = Gear C

The \ gear \ ratio = \dfrac{The \ number \ of \ teeth \ on \ the \ driven \ gear}{The \ number \ of \ teeth \ on \ the \ driver  \ gear}

The driver gear = The input gear

Therefore, we have;

The \ gear \ ratio = \dfrac{20 \ teeth}{10 \ teeth} = 2

The gear ratio = 2

-b The \ gear \ ratio = \dfrac{The \ driver \ gear\ speed}{The \ driven \ gear\ speed}

Therefore, we have;

The \ gear \ ratio = 2 = \dfrac{60 \ rpm}{Gear\ C \,  speed}

Gear\ C \,  speed = \dfrac{60 \ rpm}{2} = 30 \ rpm

-c The output (driven) gear torque at Gear C = 150 ft.·lb.

The \ gear \ ratio = \dfrac{Driven \ gear \ torque}{Driver \ gear \ torque}

Therefore;

2 = \dfrac{150 \ ft \cdot lb}{Driver \ gear \ torque}

Driver \ gear \ torque = \dfrac{150 \ ft \cdot lb}{2} = 75 \ ft \cdot lb

The input (driver) torque at Gear A = 75 ft·lb

4. -a Given that the load is between the effort and the fulcrum, we have;

The type of lever is a second class lever

-b The distance between the load and the fulcrum = 4 feet

The distance between the effort and the fulcrum = 8 feet

We have;

100 lbs × 4 ft. = Effort × 8 ft.

∴ Effort = 100 lbs × 4 ft./(8 ft.) = 50 lbs.

The effort = 50 lbs.

4 0
3 years ago
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