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3 years ago

List the parts of a manual transmission List the parts of a typical clutch assembly?

Engineering

1 answer:

True [87]3 years ago

7 0

Answer:

Explanation: Clutch Plate.

Clutch Cover.

Clutch Bearing (Release bearing)

Release Fork (clutch fork)

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Arc blow typically occurs in steel and metals that contain iron.<br> True or false

sleet_krkn [62]

Answer: True

Explanation: Iron is magnetic

4 0

3 years ago

Read 2 more answers

A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M

Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of $300\times 100 mm^2$

Force(F)=8 MN

Therefore stress due to this Force( $\sigma$ )

$\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}$

$\sigma =266.66 \times 10^{6} Pa$

$\sigma =266.66 MPa$

Since induced stress is greater than Yield stress therefore Plastic deformation occurs

8 0

3 years ago

(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced

Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

See attached image file

5 0

3 years ago

Select three functions of catalysts.

Korolek [52]

Answer: speed up food processing

speed up plant growth

Increase fuel efficiency

Explanation:

A catalyst simply refers to a substance that leads to an increase in the reaction rate when it's added to a substance. When the activation energy is reduced by catalysts, this.hwlpa on the speeding up of a reaction.

Therefore,the functions of catalysts include speed up food processing, speeding up plant growth and increase fuel efficiency

5 0

3 years ago

Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in

Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time = ∑ $(\frac{Clock cyles}{Clock rate})$

Clock cycles can be determined using following formula

Clock cycles = ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

Clock cycles = ( 50 x $10^{6}$ x 1) + ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

= $\frac{512(10^{6}) }{2(10^{9}) }$

Initial execution Time = 256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

( $\frac{Clockcycles}{2}$ )= ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

$CPI_{FP improved}$ x No. FP instructions = ( $\frac{Clockcycles}{2}$ ) -[ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = ( $\frac{512(10)^{6} }{2}$ ) - [ ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = - 206 x $10^{6}$

$CPI_{FP improved}$ = - 206 x $10^{6}$ / 50 x $10^{6}$

$CPI_{FP improved}$ = - 4.12 < 0

3 0

4 years ago