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2 years ago

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Two technicians are discussing torsion bars. Technician A says that many torsion bars are adjustable to allow for ride height ad

justment. Technician B says that torsion bars are usually marked left and right and should not be switched side to side. Which technician is correct

1 answer:

sergij07 [2.7K]2 years ago

7 0

Answer: try asking googol or sirl

Explanation:

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A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T

Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

Read more about Net Radiation Heat Transfer at; brainly.com/question/14148915

#SPJ1

8 0

2 years ago

It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less

solniwko [45]

Answer:

Acef

Explanation:

Edginuity 2021

5 0

3 years ago

A pressure gage and a manometer are connected to a compressed air tank to measure its pressure. If the reading on the pressure g

mixer [17]

Answer:

h=1.122652m

Explanation:

Assuming density of air = 1.2kg/m³

the differential pressure is given by:

$h^{i} =h(\frac{density of manometer}{density of flowing air}-1)\\h^{i} =h(\frac{1000}{1.2}-1)\\ h^{i}=832.33h...(1)\\\\but\\ h^{i} =\frac{change in pressure}{air density*g} \\\\h^{i} =\frac{11*10^3}{1.2*9.81}\\\\h^{i}= 934.42...(2)\\\\equating, \\\\934.42=832.33h\\\\h=1.122652m$

7 0

4 years ago

Why dont some vehicles have power steering

daser333 [38]

Answer:

Because older cars didn't really have much power steering, and parking or even turning at low speeds could require a lot of strength to move the steering wheel on a big, heavy car. This moves the rack to the right or left, which turns the vehicle's wheels. Without power assist, you'd need grunt force to turn the pinion and move the rack.

Explanation:

4 0

4 years ago

A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reserv

Dahasolnce [82]

This question is incomplete, the complete question is;

A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameter and is 65 ft long.

The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft.

Please estimate: The required brake horsepower.

Answer:

The required brake horsepower is 1400.08

Explanation:

Given the data in the question;

Power required to drive the pump can be determined using the formula;

P = r $_w$ QH / η₀(0.745)

given that; centrifugal pump is used to extract water from a reservoir at 14,000 gal/min.

Q = 14,000 gal/min = ( 14,000 × 0.00006309 )m³/sec = 0.883 m³/sec

the head rise across the pump is 320 ft,

H = 320 ft = ( 320 × 0.3048 )m = 97.536 m

the efficiency η₀ = 81% = 0.81

r $_w$ = 9.81 kN/m³

so we substitute our values into the formula

P = [ 9.81 × 0.883 × 97.536 ] / 0.81(0.745)

P = 844.87926528 / 0.60345

P = 1400.08 HP

Therefore, The required brake horsepower is 1400.08

5 0

3 years ago