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insens350 [35]
3 years ago
7

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

Engineering
1 answer:
erik [133]3 years ago
7 0

Answer:

your answer would be B-(Clutch-Disk)

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Ok this really isn’t a question but I need help, I’m wondering if a Samsung galaxy 9 is a good phone
Savatey [412]

Answer:

yes it is

Explanation:

it's an extremely good phone in my opinion

7 0
3 years ago
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What is the acceleration of a car that has a velocity of 20 m/s, and
ella [17]

Acceleration of Car = 10 ms⁻²

Explanation:

Step 1:

The basic formula of acceleration is a = (v-u)/t  ms⁻²

where, v- final velocity

            u- initial velocity

             t= time taken

Step 2:

Here v = 70 ms⁻¹

        u = 50  ms⁻¹

         t = 5 s

∴ a = ( 70 - 20)/5

a = 10 ms⁻²

7 0
4 years ago
A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and
vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium   K = 205 W/m k

heat flux = k \frac{dt}{dx}

               = 205 \frac{325 - 275}{0.02}

               = 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux = k \frac{dt}{dx}

               = 16.3 \frac{325 - 275}{0.02}

               = 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux = k \frac{dt}{dx}

               = 0.8 \times \frac{325 - 275}{0.02}

               = 2 KW/m2

6 0
3 years ago
A 600 MW coal-fired power plant has an overall thermal efficiency of 38%. It is burning coal that has a heating value of 12,000
velikii [3]

Answer:

See step by step explanations for answer.

Explanation:

600 megawatts =

568 690.272 btu / second

thermal eficiency=work done/Heat supllied

0.38=568690.272/Heat supplied

Heat supplied=1496553.35btu /s

heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s

feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day

sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input

so

the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %

the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day

5 0
3 years ago
Systematic searching is a skill that takes ________ to master.
bagirrra123 [75]

Answer: B, repetitive practice! hope this helps. :)

Explanation:

7 0
4 years ago
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