What is the heights part of Maine?

It is an important part of the differential maintenance which purpose is to make smoother the differential operation by lubricat

masha68 [24]

Answer:

lubrication

Explanation:

MARK ME BRAINLEIST

4 0

2 years ago

Two Technicians are discussing shock absorbers. Technician A says most shock absorbers help support vehicle weight. Technician B

yKpoI14uk [10]

Answer:

Both Technician A and B

Explanation:

In a vehicle suspension system, a shock absorber has the functions such as; limiting the movement of the vehicle, stabilizing the ride, stabilizing the tires of the vehicle, minimizing wear and tear in the vehicle and decreasing the overall suspension tear. Some of the ways of maintaining shock absorbers is to inspect the strut for leaks.Shock oil sometimes leaks to cover the seals with a transparent liquid.

6 0

3 years ago

A tank with some water in it begins to drain. The function v ( t ) = 46 − 3.5 t determines the volume of the water in the tank (

olchik [2.2K]

Answer with Explanation:

Part a)

The volume of water in the tank as a function of time is plotted in the below attached figure.

The vertical intercept of the graph is 46.

Part b)

The vertical intercept represents the volume of water that is initially present in the tank before draining begins.

Part c)

To find the time required to completely drain the tank we calculate the volume of the water in the tank to zero.

$0=46-3.5t\\\\3.5=46\\\\\therefore t=\frac{46}{3.5}=13.143minutes$

Part d)

The horizontal intercept represents the time it takes to empty the tank which as calculated above is 13.143 minutes.

7 0

3 years ago

Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type co

Softa [21]

Answer:

I am Providing Answer in C Language Program.

Explanation:

Please find attachment regarding code of taking two numbers input and adding them.

I would like to recommend you please use software which supports C language.

#include <stdio.h>

int main () {

int a, b, sum;

printf ("\ nEnter two no:");

scanf ("% d% d", & d, & e);

sum1 = d + e;

printf ("Sum:% d", sum1);

return (0);

}

4 0

3 years ago

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago