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a_sh-v [17]
3 years ago
15

When adjusting the valve clearances on the engine, Technician A says the valves are closed when the valve lifter or follower is

on the inner base circle of the camshaft lobe. Technician B says that valves are always adjusted when closed while the lifter or follower is on the camshaft lobe nose. Who is correct?
A. Technician A only
B. Technician B only
C. Both A and B
D. Neither A nor B
Physics
1 answer:
andre [41]3 years ago
7 0

Answer:

A. Technician A only

Explanation:

The camshaft opening ramp is what opens the valve, it starts at the basement of the inner circle and closes at the nose while the camshaft closing ramp allows for closing a valve starts at the nose and closes at the inner base circle. Valve clearance on an engine are performed when closing a valve, and is completed when the valve lifter or follower is on the inner base circle of the camshaft lobe.

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Zielflug [23.3K]

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

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3 years ago
What is the difference between regional metamorphism and contact metamorphism?
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Contact metamorphism<span> is a type of </span>metamorphism<span> where rock minerals and texture are changed, mainly by heat, due to </span>contact<span> with magma. </span>Regional metamorphism<span> is a type of </span>metamorphism<span> where rock minerals and texture are changed by heat and pressure over a wide area or region.</span>
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3 years ago
An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
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Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

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Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

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\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

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