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3 years ago

11

A digital speedometer constantly reads zero mph. Technician A says the problem may be the vehicle speed sensor. Technician B say

s the problem may be the throttle position sensor. Who is correct

1 answer:

uysha [10]3 years ago

4 0

Answer:

Technician A

Explanation:

The vehicle speed sensor is located at the right side of the transmission near the output shaft, it is also known as the output speed sensor.it is what tells your vehicle control unit how fast it is going. The vehicle speedometer and odometer rely on the information from this sensor.

With a faulty speed sensor ,you might not be able to know how fast your car is going.

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What happens to the demand for butter, if the price of margarine increases? For Economics.

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Supply and demand goes up .

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3 years ago

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What is the term used to label the energy levels of electrons?

mel-nik [20]

Electron volts...........

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3 years ago

Voltage will be induced in a wire loop when a magnetic field within that loopa. is at right angles to the electric field b. chan

mixas84 [53]

Answer:

changes

Explanation:

According to Michael Faraday, voltage is induced on a metallic conductor when the magnetic field changes. The principle is known as the principle of electromagnetic induction. The voltage induced on the metallic conductor is referred to as induced emf.

The magnitude if induced emf depends on the rate of change of the magnetic flux.

3 0

3 years ago

Light in air enter crown glass (n = 1.52) at an angle of 35 degrees. What is the refracted angle?

Dovator [93]

Answer:

22.17 degree

Explanation:

n = 1.52

Angle of incidence, i = 35 degree

Let the angle of refraction is r.

use the Snell's law

n = Sin i / Sin r

Sin r = Sin i / n = Sin 35 / 1 .52

Sin r = 0.37735

r = 22.17 degree

Thus, the ray is refracted at an angle of 22.17 degree.

8 0

3 years ago

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. I

oksano4ka [1.4K]

Answer:

The deviation in path is $4.39 \times 10^{-3}$

Explanation:

Given:

Velocity $v = 1 \times 10^{6}$ $\frac{m}{s}$

Electric field $E = 500$ $\frac{V}{m}$

Distance $x = 1 \times 10^{-2}$ m

Mass of electron $m = 9.1 \times 10^{-31}$ kg

Charge of electron $q = 1.6 \times 10^{-19}$ C

Time taken to travel distance,

$t = \frac{x}{v}$

$t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }$

$t = 10^{-8}$ sec

Acceleration is given by,

$F = qE$

$ma = qE$

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }$

$a = 8.77 \times 10^{13}$ $\frac{m}{s^{2} }$

For finding the distance, we use kinematics equations.

$y = vt + \frac{1}{2} at^{2}$

Where $v = 0$ because here initial velocity zero

$y = \frac{1}{2} at^{2}$

$y = \frac{1}{2} \times 8.77 \times 10^{13 } \times (10^{-2} )^{2}$

$y = 4.39 \times 10^{-3}$ m

Therefore, the deviation in path is $4.39 \times 10^{-3}$

6 0

2 years ago