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ella [17]
1 year ago
15

technician a says an out-of-round drum can cause a pulsating brake pedal. technician b says an out-of-round drum can cause the b

rakes to grab. who is correct?
Engineering
1 answer:
denis-greek [22]1 year ago
8 0

According to the question of the pulsating brake pedal, both A and B are correct.

What causes brake pulsation?

Brake pulsation is mainly caused by warped rotors/brake discs. Excessive hard braking or quick stops, which can significantly overheat the discs, are the primary causes of deformed rotors. When the discs overheat, the composition of the metal disc material changes, resulting in imperfections in the metal's surface. Hotspots are noticeable irregularities. They appear as discoloured areas of the disc material, which are often bluish or blackish in appearance. The brake pedal is the pedal which you press with your foot to slow or stop a vehicle. When the driver presses the brake pedal, the system automatically delivers the appropriate pressure required to prevent colliding with the vehicle in front.

To learn more about brake pulsation
brainly.com/question/28779956

#SPj4

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Q.14
Kay [80]

Answer:

A. optical isolation

Explanation:

well I can't really give a good explanation because I also saw the same question in my exams and option A was the correct answer

6 0
3 years ago
Urgent please help me...
Pepsi [2]
Scuze , am incercat , dat chiar nu stiu. Prefer sa nu raspund si sa scriu asta decat sa raspund gresit.❤️
7 0
3 years ago
A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t
bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

a=450\ mm/s^2

Explanation:

Given that

s = 15t^3 - 3t\ mm

a)

When t= 2 s

s = 15t^3 - 3t\ mm

s = 15\times 2^3 - 3\times 2\ mm

s= 114 mm

At t= 4 s

s = 15t^3 - 3t\ mm

s = 15\times 4^3- 3\times 4\ mm

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

b)

We know that velocity V

V=\dfrac{ds}{dt}

\dfrac{ds}{dt}=45t^2-3

At t=  5 s

V=45t^2-3

V=45\times 5^2-3

V=1122 mm/s

We know that acceleration a

a=\dfrac{d^2s}{dt^2}

\dfrac{d^2s}{dt^2}=90t

a= 90 t

a = 90 x 5

a=450\ mm/s^2

4 0
3 years ago
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

  diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0
2 years ago
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Pavlova-9 [17]

Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

8 0
3 years ago
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