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insens350 [35]
2 years ago
7

Which part does NOT rotate when the engine is running and the clutch pedal is depressed?

Engineering
1 answer:
erik [133]2 years ago
7 0

Answer:

your answer would be B-(Clutch-Disk)

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A spherical metal ball of radius r_0 is heated in an oven to a temperature of T_1 throughout and is then taken out of the oven a
telo118 [61]

Answer:

yes it is u are right

Explanation:

do be secure

7 0
2 years ago
Our goal is to design a traffic-light controller with the following properties; it lights up the green light (output G) for 15 s
nataly862011 [7]

Answer:

There will be three (3) states that will be needed in-order to implement the FSM controller.

Explanation:

FSM stands for  Finite-State Machine. This is an example of a model used in  mathematical computation. The FSM is an abstract machine that can be in exactly one of a finite number of states at any given time.

3 0
3 years ago
The price of a single item within a group of items is
Nady [450]

Answer:

Answer to this question is option D i.e. unit price.

Explanation:

The unit price of the item can be understood as the price of a single product or one single commodity which forms a part of a group of items. When only one unit is to be sold then here comes the importance of 'unit price.' This is generally helpful in the retail sector where the products are bought in bulk after calculating the per-unit price of each commodity in that particular bulk.

7 0
3 years ago
List the parts of a manual transmission <br><br> List the parts of a typical clutch assembly?
True [87]

Answer:

Explanation: Clutch Plate.

Clutch Cover.

Clutch Bearing (Release bearing)

Release Fork (clutch fork)

7 0
2 years ago
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

7 0
3 years ago
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