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baherus [9]
3 years ago
10

Technician A says that torque-to-yield head bolts are special bolts that are reusable. Technician A says that main bearing clear

ance can be calculated by knowing the inner diameter of the main bearing and the outer diameter of the main journal. Technician B says that rod bearing and main bearing clearance can be measured with Plastigauge. Who is correct?
A) Technician A only
B) Technician B only
C) Both A and B
D) Neither A nor B
Physics
1 answer:
Blizzard [7]3 years ago
7 0

Answer: D

Neither A nor B

Explanation:

In order to check the clearances for rod and main bearings, you need a set of micrometers and a dial-bore gauge

Measuring the inside diameter of a main or rod bearing will require a dial bore gauge. The best ones to use are accurate down to 0.0001-inch.

So, both technician A and B are incorrect

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Determine the amount of potential energy of a 5N book that is 1.5m high on a shelf.
Alex73 [517]

Answer:

Potential energy of book = 7.5 J

Explanation:

Given:

Weight of book = 5 N

Height of shelf = 1.5 meter

Find:

Potential energy of book

Computation:

Weight = Mass x Acceleration of gravity

Mass x Acceleration of gravity = 5 N

Potential energy = Mass x Acceleration of gravity x Height

Potential energy of book = Mass x Acceleration of gravity x Height

We know that;

Mass x Acceleration of gravity = 5 N

So,

Potential energy of book = 5 x 1.5

Potential energy of book = 7.5 J

3 0
3 years ago
Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the
masya89 [10]
The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2


8 0
4 years ago
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
Read 2 more answers
_____ is very corrosive and can cause rusting so metal tanks shouldn't be used
bagirrra123 [75]

Answer:

Iron

Explanation:

5 0
3 years ago
There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

7 0
3 years ago
Read 2 more answers
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