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IceJOKER [234]
3 years ago
13

Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as

necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors
Physics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

the number of carousel conveyors that an operator can operate without any idle time is 5

Explanation:

Given the data in the question;

first we express the equation for number of carousel conveyors that can be operated by an operator;

n' = \frac{(a + t)}{( a + b)}

where a is the concurrent activity time ( 0.25 minute )

b is the independent operator activity time

t is the independent machine activity time( 1 )

Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.

So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item

so

we substitute

0.25min for a, 1 for t and 0min for b

n' = \frac{(0.25min + 1min)}{( 0.25min+ 0 min)}

n' = 1.25 min / 0.25

n' - 5

Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5

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Answer:

Correct option A.

The net force exerted by this loop on the straight wire with the current is directed TOWARDS THE LOOP

Explanation:

The magnetic field exerts a force on a current-carrying wire in a direction given by the right-hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire since typical currents consist of very large numbers of moving charges.

Given that,

The wire's current is directed towards the right of the page.

The rectangular loops carry current in a clockwise direction.

Since the 'dot' field is increasing hence the induced magnetic field is 'cross', i.e. into the page and by the right-hand rule, the induced current is clockwise.

Then the magnetic field is into the page.

Since was known that

F= iL×B

Note the current is through the wire. Then, the length is in direction of the current.

Note: this equation gives the magnetic force that acts on a length L of a straight wire carrying a current (i) and immersed in a uniform magnetic field (B), that is perpendicular to the wire.

So, the magnetic field is always perpendicular to the current.

So using right hand rule,

F = i(L×B)

The length is to the right i.e. +x direction and the Magnetic field is perpendicular to the plane, i.e. in the +z direction

F = i (L•i × B•k)

F = iLB (i×k)

F = iLB•(-j)

F = -iLB•j

Then, the force is in the negative y-direction i.e. towards the loop.

8 0
4 years ago
What might be done to prevent acid rain damage to objects made from metal and carbonate, such as limestone? Identify a metal obj
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Answer:  Bronze is a metal that can be damaged in acid rain. It is harsh to metals by almost burning them and may create rust. Everbrite is a clear, protective coating that will seal metal and bronze.

Explanation:

I know this is correct i have done all the research

7 0
3 years ago
What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p
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Answer:

A. Volume is unchanged

Explanation:

P_{i} = initial pressure of the gas = 722 torr = 96258.7 pa

P_{f} = final pressure of the gas = 0.950 atm = 96258.75 pa

T_{i} = initial temperature = 32 °F = 272.15 K

T_{f} = final temperature = 273 K

V_{i} = initial volume

V_{f} = final volume

Using the Equation

\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}

Inserting the values

\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}

V_{f} = (1.00312) V_{i}

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4 0
3 years ago
Anissa slides down a playground slide sloped at 25o. The coefficient of kinetic friction between Anissa and the slide is 0.15. I
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Answer:5.62 m/s

Explanation:

Given

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mass of Anissa's m=55 kg

time interval t=2 s

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a=g\sin \theta -\mu _kg\cos \theta

a=9.8\sin (25)-0.15\cdot 9.8\cos (25)

a=4.14-1.33=2.81 m/s^2

v=u+at

where v=Final Velocity

u=Initial Velocity

a=acceleration of system

t=time

v=0+2.81\times 2

v=5.62 m/s

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