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Sloan [31]
3 years ago
8

Two technicians are discussing cylinder- testing. Technician A says that when testing the power level of a specific cylinder usi

ng a Power balance test, it's best to ground it's secondart current. Technician B says that it's best to disconnect one spark plugs wire at a time to test cylinder power levels. Who is correct?​
Engineering
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

The answer is "Both Technician A and Technician B".

Explanation:

The cylinder Testing is intended to assess locomotive inconsistency in CNS rodents, for example, whenever the animal moves within a transparent plastic tube, its preliminary activity is registered as it rises against the stadium wall.

In the given question both technicians are correct because both are reliable ways to check cylinders and the influence of the belief if every pathway has many more advantages than each other.

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Match the word with the definition:
aksik [14]

1. Renewable Resources  = (Renewable means you can keep making it) =  resources that can be replenished (such as trees)

2. Nonrenewable Resources  =  ( Nonrenewable means it can't be made once it is used up) = resources that are gone once they are used (such as fossil fuels)

3. Producer  =  ( produces something) = person who makes goods or provides services

4. Consumer  = ( uses something) =   person whose wants are satisfied by using goods and services

5. Allocate  = ( put someplace) =   distribute

6. Choice =  option

7 0
3 years ago
Read 2 more answers
3. Suppose you work for this company that evaluates Boolean circuits in exponential time. Since you are very smart, your manager
hammer [34]

Answer:

Ergr5

Explanation:

3 0
3 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the
muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, \L_e = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

P =\frac{\pi^2EI}{L_e^2}

where, I is the moment of inertia

on substituting the respective values, we get

10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}

or

I = 13.13 in⁴

also for circular cross-section

I = \frac{\pi}{64}\times d^4

thus,

13.13 = \frac{\pi}{64}\times d^4

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0
3 years ago
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