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slamgirl [31]
2 years ago
7

Technician a says that a defective spark plug wire can cause an engine miss. technician b says that a defective spark plug can c

ause an engine miss. which technician is correct?
Physics
1 answer:
larisa86 [58]2 years ago
7 0

Technician a says that a defective spark plug wire can cause an engine misfire. Technician b says that a defective spark plug can cause an engine misfire. In this case, both technicians are correct.

Technician A is correct, because, Bad spark plug wires can interfere with the electrical current flowing to the engine, making it difficult for the engine to complete the combustion cycle. An incomplete combustion cycle can cause an engine to misfire, also

Technician B is correct because, Dirty spark plugs can cause misfires as well as burned engine oil can prevent the plugs from creating the sparks needed to ignite the fuel after it goes into the cylinder. Old spark plugs can simply break and fail to produce a spark. A broken spark plug is a simple fix, just replace it.

To learn more about Engine misfire, here

brainly.com/question/28204007

#SPJ4

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Rose has been working as a receptionist for 20 years. One day, she receives a check in the mail for $750,000—an inheritance from
sergey [27]
The just-world phenomenon is the belief that everything that happens to an individual is due to the individual's actions; in other words, all good and all bad that an individual encounters in the world is deserved by that person. This leads to a victim being blamed with the logic that "they had it coming" and someone who encounters good fortune being praised with "they earned it". Therefore, in this scenario, people will assume that Rose's inheritance is well deserved.<span />
8 0
3 years ago
Read 2 more answers
A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,
borishaifa [10]

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

5 0
3 years ago
A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0
3 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0
3 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
4 years ago
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