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

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' =

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' =

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

Independent variable: the best method to get rid of them.

Dependent variable: washing with soap and water.

Hypothesis: Organic oils

Control group: water

Answer:

Step-by-step explanation:

D.
Heat each cube to the same temperature, place each cube into different containers with 500 grams of water at the same temperature, and measure the temperature of the water.

Answer:

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

The question specifies the diameter of the screw, therefore the IMA of this screw is 0.812? / 0.318 = 8.02