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2 years ago

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6. After removing the vacuum hose from the fuel-pressure regulator, you see gasoline has dripped out of the hose. Technician A s

ays that the fuel pressure regulator is defective. Technician B says that one or more of the injectors may be defective, causing the fuel to get into the hose. Who is correct?

1 answer:

lubasha [3.4K]2 years ago

7 0

Answer:

Technician B is correct.

Explanation:

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A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000 $\frac{ft lbm}{s^{2} }$

We know that moment of inertia is given as

$I = \frac{bt^{3} }{12}$

$I = \frac{5 (1.5^{3} )}{12}$

I = 1.40625 $in^{4}$

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = $\frac{PL^{3} }{48EI}$

$0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}$

E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0

2 years ago

In the engineering design process, testing is the least important step.<br> True<br> False

Wittaler [7]

Answer:

false

Explanation:

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2 years ago

Read 2 more answers

Please look at the picture for question.

Vilka [71]

Answer:

more than 10 joules jjsjdnjdfjnfjfnjfnnfndnndjnfjgngjntjfjmrmdkdmjd

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1 year ago

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A steam turbine has isentropic efficiency of 0.8. Isentropically, it is supposed to deliver work of 100 kW. What is the actual w

trapecia [35]

Answer:

80 kW; 11 kW; 8 kW; 0.6

Explanation:

Part 1

Isentropic turbine efficiency:

$\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s}$

$W_{real} = \eta_t*W_s$

$W_{real} = 0.8*100 kW$

$W_{real} = 80 kW$

Part 2

Coefficient of performance COP is defined by:

$COP = \frac{Q_{out}}{W}$

$Q_{out} = W*COP$

$Q_{out} = 5 kW*2.2$

$Q_{out} = 11 kW$

Part 3

(a)

Energy balance for a refrigeration cycle gives:

$Q_{in} + W = Q_{out}$

$3 kW + 5 kW = Q_{out}$

$8 kW = Q_{out}$

(b)

$COP = \frac{Q_{in}}{W}$

$COP = \frac{3 kW}{5 kW}$

$COP = 0.6$

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2 years ago

A hypothetical accumulator processor uses one of the following 16-bit instruction formats, depending on the instruction.(a) (10

Jlenok [28]

Answer:

Every part has been answered with complete detail. See the pictures.

Explanation:

See the pictures for detailed answer.

7 0

3 years ago