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Helga [31]
3 years ago
12

While discussing the effects of loose wheel bearings: Technician A says that vibration may occur while driving. Technician B say

s that the bearing may make noise. Who is correct? a. Technician A b. Neither A nor B c. Technician B d. Both A and B
Physics
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

Option "D" is the correct answer i.e. Both Technicians A & B

Explanation:

Dear,

Loose wheel bearings can be caused on any machine.

The Following can be the reasons :

  • the machine or motor is installed improperly
  • the rotor(rotating part) of machine is not balanced
  • the cover get deteriorated with time
  • Bad manufacturer

So, when the bearings will be loose then the following can happen:

  1. It will produce vibration in machine which can be judged by a vibration pen eg. SKF vibration pen
  2. The vibration will also result in noise. You can check by
  3. The machine winding if it is motor will burn-out.
  4. Bush needs to be installed if the bearing runs loose.

Therefore,  "D" is right.

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Reflection

Explanation:

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3 years ago
a body of mass 5kg falls from height of 10m above the ground what kinetic energy of the body before it strike the ground
Snowcat [4.5K]
Gravitational Potential Energy (GPE) before fall = Kinetic energy on impact
GPE = mgh
GPE = 5kg x 9.8m/s^2 x 10m
GPE = 490 J
Kinetic Energy = 490 J

(This is assuming that gravitation field strength (g) is 9.8m/s^2, sometimes       you may round this to 10m/s^2, therefore making the final result: 
  Kinetic energy = 500 J)
4 0
4 years ago
Which of the following is a form of kinetic energy?
aalyn [17]

Answer:

A) Sound Energy

Explanation:

Electrical and nuclear energy are examples of potential energy

6 0
4 years ago
Read 2 more answers
The contraption below has the following
AnnyKZ [126]

Answer:

117.72 N

Explanation:

The given parameters are;

The mass m₁ = 2.0 × 10³ kg

The mass m₂ = 4.4 × 10² kg

The mass of the man, m₃ = 6.0 × 10 kg

The condition of the interaction of the surfaces = Frictionless surfaces

The

The tension in the string = The downward force = The weight of (m₂ + m₃) = (m₂ + m₃) × g

Let <em>a</em> represent the acceleration of the connected masses due to the weight of m₂, and m₃, we have;

(m₁ + m₂ + m₃) × a = (m₂ + m₃) × g

∴ a = (m₂ + m₃) × g/(m₁ + m₂ + m₃)

Which gives;

a = (4.4 × 10²+ 6.0 × 10) × 9.81/(2.0 × 10³+ 4.4 × 10²+ 6.0 × 10) = 1.962

The downward acceleration, a = 1.962 m/s²

The apparent weight of the man = The mass of the man, m₃ × The acceleration, <em>a</em>

∴ The apparent weight of the man = 6.0×10 kg ×1.962 m/s² = 117.72 N

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%
Verdich [7]

Answer:

1.7 × 10^1^0

Explanation:

Knowing that, the volume of the sphere is given by, v=\frac{4}{3}\pi ^3

Thus, the fractional change in volume is given by,

=3 × \frac{0.02}{100}=\frac{0.06}{100}

Pressure at the bottom of the sea is,

Δp =pgh = 10^3 × 10 × 10^3=10^7 pa.

Knowing that,

Bulk modulus: \frac{10^7 * 100}{0.06}=\frac{10^9}{6*10^-^2}=\frac{10^1^1}{6}

B =\frac{10}{6}*10^1^0

B = 1.7*10^1^0N/M^2

Answer = 1.7 × 10^1^0

[RevyBreeze]

5 0
2 years ago
Read 2 more answers
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