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Dafna1 [17]
3 years ago
9

19. What part of the engine is shown at point "A" in the above figure?

Engineering
2 answers:
stepan [7]3 years ago
6 0

Answer:

C. Four bolt main

Explanation:

I think.. :)

erik [133]3 years ago
4 0
The answer is c I took the quiz my self
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As a general rule of thumb, the ratio of the rate of etch-product formation to the flow rate of etch gas should be greater than
g100num [7]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

8 0
3 years ago
An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box
True [87]

Answer:

Rate of Entropy =210.14 J/K-s

Explanation:

given data:

power delivered to input = 350 hp

power delivered to output = 250 hp

temperature of surface = 180°F

rate of entropy is given as

Rate\  of\ entropy  = \frac{Rate\ of \ heat\  released}{Temperature}

T = 180°F = 82°C = 355 K

Rate of heat = (350 - 250) hp = 100 hp = 74600 W

Rate of Entropy= \frac{74600}{355} = 210.14 J/K-s

8 0
3 years ago
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
3 years ago
Why must you compress the air-fuel mixture inside the combustion chamber?
8090 [49]

Answer:

all above bc heat is heat fuel is fuel

3 0
3 years ago
Read 2 more answers
Pt 1. Amanda lifts a crate into the back of her truck. She exerts a force of 200N distance of 1.2m. Calculate the amount of work
melomori [17]

Answer (Part 1):

620 joules

Explanation:

Work is done when force is applied on an object over a distance. It is measured in joules.

<u>Formula</u>

Thus, Work = Force x distance

Since, Work = ? (unknown value)

<u>Given:</u>

Force = 200N

Distance = 1.2 meters

<u>So,</u>

Work = 200N x 3.1 meters

= 620 joules

Thus, Amanda did a work of 620 joules on the crate.

Answer (Part 2):

240.25 joules

Explanation:

Work is done when force is applied on an object over a distance. It is measured in joules.

<u>Formula</u>

Thus, Work = Force x distance

Since, Work = ? (unknown value)

<u>Given:</u>

Force = 77.5N

Distance = 3.1 meters

<u>So,</u>

Work = 77.5N x 3.1 meters

= 240.25 joules

Thus, Amanda did a work of 240.25 joules on the crate.

Hope this helps! :)

7 0
3 years ago
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