All categories

3 years ago

When lubricating parts that do not have a grease relief port, inject grease until

Engineering

1 answer:

IRISSAK [1]3 years ago

7 0

Answer:

C. The boot swells slightly

Explanation:

Friction can be defined as a force that opposes the motion of an object in relation to another.

This ultimately implies that, friction is the opposition to the relative motion of two objects in contact. Therefore, lubrication (lubricants) are used to prevent friction between moving parts.

Generally, grease relief port are created by manufacturers of mechanical parts so as to prevent over-lubrication.

However, when lubricating parts that do not have a grease relief port, inject grease until the boot swells slightly so as to maintain the seal at the ball joints.

You might be interested in

A high-voltage direct-current (dc) transmission line between Celilo, Oregon and Sylmar, California is 845 mi in length. The line

Gennadij [26K]

Answer:

<u>Total Mass of Conductors = 19913661.3 kg</u>

Explanation:

The total length is given as:

Total Length = 845 mi

Converting it to ft:

Total length = (845 mi)(5280 ft/1 mi)

Total length = 4,461,600 ft

The weight of one conductor per unit length is:

Mass of one conductor per unit length = 2460 lb/1000 ft

Mass of one conductor per unit length = 2.46 lb/ft

Since there are 4 conductors, therefore, total weight of all conductors per unit length will be:

Mass of all four conductors per unit length = 4 x 2.46 lb/ft

Mass of all four conductors per unit length = (9.84 lb/ft)(0.453592 kg/1 lb)

Mass of all four conductors per unit length = 4.463 kg/ft

Therefore, total mass of conductors in the line will be:

Total Mass of Conductors = (Mass of all four conductors per unit length)(Total Length)

Total Mass of Conductors = (4.463 kg/ft)(4,461,600 ft)

<u>Total Mass of Conductors = 19913661.3 kg</u>

4 0

3 years ago

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He

Lyrx [107]

<h3>Answer:</h3>

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He works with the <u>Workers</u> , which consists of members who are building code officials and building safety professionals.

8 0

2 years ago

A motor cycle is moving up an incline of 1 in 30 at a speed of 80 km/h,and then suddenly the engine shuts down.The tractive resi

Ad libitum [116K]

Explanation:

option iii is the right answer.

4 0

3 years ago

A student is building a circuit which material should she use for the wires and why?

Nataliya [291]

Answer:

i think it is D tell me if its wrong

Explanation:

6 0

3 years ago

Read 2 more answers

A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago