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3 years ago

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

1 answer:

7 0

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.Make sure the cables are not damaged and in good condition.Keep the cables out of the work area.Wear proper personal protective equipment:eye, ear and face protection,metatarsal safety boots, where required,respiratory protection may be required, depending on the work.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.Wheels are made only for grinding certain items. Do not grind rough forgings on a small precision grinding wheel. is that enough

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The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers

increase of green house gases is warming the oceans and melting sea ice at the poles. This causes changes in climate. Name three

Ahat [919]

core, surface atmosphere

4 0

4 years ago

Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?

kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time = 4s

Momentum = 1000kgm/s

Unknown:

Force = ?

Solution:

To solve this problem, we use Newton's second law of motion;

Ft = Momentum

F is the force

t is the time

So;

F x 4 = 1000kgm/s

F = 250N

6 0

3 years ago

A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o

umka2103 [35]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

$F = qv \times B = qvBsin(\theta)$ (1)

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

$v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s$

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

$K = \frac{1}{2}mv^{2}$

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

$K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV$