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Paha777 [63]
3 years ago
13

A "chirping" noise is heard while the vehicle is moving forward, but stops when the brakes are applied. Technician A says that t

he noise is likely caused by the disc brake pad wear sensors. Technician B says the noise is likely a wheel bearing because the noise stops when the brakes are applied. Which technician is correct?
A. Technician A only.
B. Technician B only.
C. Both Technicians A and B.
D. Neither Technicians A and B.
Physics
2 answers:
Alina [70]3 years ago
7 0

Answer:

A. Technician A only

Explanation:

A "chirping" noise is heard while the vehicle is moving forward, but stops when the brakes are applied. Technician A says that the noise is likely caused by the disc brake pad wear sensors. Technician B says the noise is likely a wheel bearing because the noise stops when the brakes are applied. In the given scenario <em>technician A is correct.</em>

alexdok [17]3 years ago
4 0

Answer:

Technician A.

Explanation:

A Brake wear indicator is used to warn the user and/or owner of a vehicle that the brake pad is in need of replacement. By rubbing against the discs, these make a loud screeching sound, providing an audio clue to the driver that the pads have reached their maximum wear limits.

The classic sounds of a bad wheel bearing are cyclic chirping, squealing and/or growling noise. You can also tell that the sound is related to wheel bearings if it changes in proportion to vehicle speed. The sound can get worse with every turn, or it can disappear momentarily.

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pychu [463]
Velocity = displacement / time
v = 6 / 3/4
v = 6*4/3
v = 24/3
v = 8 m/s toward north

In short, Your Answer would be 8 m/s north

Hope this helps!
8 0
4 years ago
Read 2 more answers
An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle
shtirl [24]

Answer:

-9.45\times10^{-16} \text{ J}

Explanation:

According to Coulomb's law, the force of attraction between two point charges, q_1 and q_2, separated by a distance d is given by

F = k\dfrac{q_1q_2}{d^2}

k is a constant with a value of 9\times10^9\text{ F/m}.

When we substitute the values from the question,

F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

W = F\times r

W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}

This is negative because work is done on the electron, not by it.

3 0
3 years ago
At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she
insens350 [35]

Answer:

1.34 * 10^{3}m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, v_{x} = 990m/s

Surface charge density, σ = 2.34 x 10^{-9} C/m^{2}

We need to calculate the speed at time, t = 7.0 * 10^{-8}s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed v_{x} on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity v_{y} in the y axis.

Hence, the resultant velocity of the proton is given by:

v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}

v_{x} = 990m/s from the question. We need to find v_{y} and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }

E =  132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }

a = 1.3 * 10^{10} m/s^{2}

Therefore, at time, t = 7.0 * 10^{-8}, we can use one of the equations of linear motion to find the velocity in the y axis:

a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at

v_{y} = 0 + (1.3 * 10^{10} * 7.0 * 10^{-8})

v_{y} = 910 m/s

∴ v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}

v = \sqrt{990^{2} + 910^{2} }

v = \sqrt{1808200}

v = 1344.69 m/s = 1.34 * 10^{3}m/s

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Which of the following is something you could do to reduce your impact on the environment?
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Answer:

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