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Mrac [35]
3 years ago
9

In order to be sure a grinder used in production is operating properly, what feature of the grinder should be checked daily?

Physics
1 answer:
Luba_88 [7]3 years ago
7 0
<span>Fasten pedestal and bench grinders on a solid surface securely.Ensure all the guards are in place and secure before using a grinder.Adjust tool rests to within 3 mm (1/8 in.) of wheels. Never adjust rests while wheels are moving. Work rest height should be on horizontal centre line of the machine spindle.Maintain 6 mm (1/4 in.) wheel exposure with a tongue guard or a movable guard.Check that wheels have blotters on each side.Check the wheel fits properly to the spindle when mounting. If it is loose, get another wheel.Tighten the nuts before you turn the grinder on.Before you plug in the grinder, manually spin the wheel to make sure it is spinning freely.</span><span>Make sure the cables are not damaged and in good condition.Keep the cables out of the work area.<span>Wear proper personal protective equipment:<span>eye, ear and face protection,metatarsal safety boots, where required,respiratory protection may be required, depending on the work.</span></span>Wear gloves only where necessary and if there is no risk of entanglement.Stand to one side of the grinder until the wheel reaches operating speed.Bring work into contact with the grinding wheel slowly and smoothly, without bumping.Apply gradual pressure to allow the wheel to warm up evenly. Use only the pressure required to complete a job.Move the work back and forth across the face of the wheel. This movement prevents grooves from forming.<span>Wheels are made only for grinding certain items. Do not grind rough forgings on a small precision grinding wheel. is that enough</span></span>
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Relate the output of energy from a heat engine to the energy put into the heat engine considering the second law of thermodynami
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<u>Answer: </u>

<em>Considering the II law of thermodynamics</em>

<em>From the figure</em>

<em>Out put of energy: </em>

Heat supplied from the source/ reservoir  (Q₁) - Heat rejected to the surroundings from the system (Q) = Q₁ - Q₂. Also known as Net work done on the system.

<em>Input of energy: </em>

Amount of heat energy supplied to the system from the source (Q₁ ).

Efficiency (H.E) = η = Output÷ Input

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                      OR η = Wnet ÷ Q₁ ;        since Wnet = (Q₁ - Q₂)


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Energy that is 'lost" to heat, has merely been converted to another form, not<br> lost forever.
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In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
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Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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