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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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1 kg lead to earth is greater attraction as mass of earth is much more than 1kg lead.

Explanation:

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In a transverse wave the particles Name
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2 years ago
A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to a
kogti [31]

Answer:

a) τ =  0.672 N m , b) θ = 150 rad , c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

       w = w₀ + α t

       α = -w₀ / t

       α = 120 / 2.5

       α = 48 rad / s²

The moment of inertia of a cylinder is

       I = ½ M R²

Let's calculate the torque

      τ = I α

      τ = ½ M R² α

      τ = ½ 2.8 0.1² 48

      τ =  0.672 N m

b) we look for the angle by kinematics

      θ = w₀ t + ½ α t2

      θ = ½ α t²

      θ = ½ 48 2.5²

      θ = 150 rad

c) work in angular movement

      W = τ θ

      W = 0.672 150

      W = 100.8 J

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3 years ago
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3 years ago
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Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generat
DiKsa [7]

Answer:

The power in this flow is 9.56\times10^{8}\ W

Explanation:

Given that,

Distance = 221 m

Power output = 680 MW

Height =150 m

Average flow rate = 650 m³/s

Suppose we need to calculate the power in this flow in watt

We need to calculate the pressure

Using formula of pressure

Pressure=\rho g h

Where, \rho= density

h = height

g = acceleration due to gravity

Put the value into the formula

Pressure=1000\times9.8\times150

Pressure=1470000\ Nm^2

We need to calculate the power

Using formula of power

P=Pressure\times flow\ rate

Put the value into the formula

P=1470000\times650

P=9.56\times10^{8}\ W

Hence, The power in this flow is 9.56\times10^{8}\ W

5 0
3 years ago
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