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BartSMP [9]
3 years ago
5

A clutch is chattering. Technician A says that uneven wear on the flywheel could be the cause. Technician B says that worn or br

oken motor mounts could be the cause. Who is correct? Group of answer choices
Engineering
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

Both Technician A and B are correct

Explanation:

Due to a contamination of the clutch disc friction surface, the clutch chatters

The contamination that causes clutch chattering includes

1) Worn or broken motor mounts

2) Bell housing bolts becoming loose

3) Clutch link damage

4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering

5) Engine or transmission oil contaminating the disc.

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List six possible valve defects that should be included in the inspection of a used valve?
olchik [2.2K]

Answer:

Valvular stenosis , Valvular prolapse , Regurgitation,

Explanation:

8 0
3 years ago
Please help <br> please i need to turn this in
grigory [225]

Answer:  101 means "introductory something". The allusion is to a college course with the course code 101, which in the American system and probably others indicates an introductory course, often with no prerequisites.

Explanation:

The name Topher means Christ Bearer and is of American origin. Topher is a name that's been used primarily by parents who are considering baby names for boys. Short form of Christopher.

I am not sure but if this is a trick question then the answer lies with his name, so this is the best I got. Hope it helps!

5 0
3 years ago
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W
cupoosta [38]

Answer:

It will not  experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:

<em>σc = KIc / (Y*√(π*a))</em>

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 0
3 years ago
A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio
Mama L [17]

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

P_{incandescent} = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

P_{fluorescent} = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

P_{saved} = P_{incandescent}  - P_{fluorescent}

P_{saved} = 750 - 180

P_{saved}  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

3 0
3 years ago
Describe the Range of a set of numbers and how it is calculated.
allochka39001 [22]

Answer:

The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

Explanation:

7 0
3 years ago
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