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muminat
3 years ago
12

In Process Costing, the journal entry to record the cost of goods transferred out from the WIP shaping department to the WIP pac

kaging department would be a debit to WIP - shaping department and a credit to WIP - packaging department. True False
Engineering
1 answer:
ddd [48]3 years ago
4 0

Answer:

Journal entry :

Date Account and explanation Debit credit

                WIP-Packing department XXX  

                WIP-Shaping department            XXX

(To record cost of goods transferred out from the WIP shaping department to the WIP packaging department )  

So above statement is False

Explanation:

You might be interested in
Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl
ludmilkaskok [199]

Answer:

Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

Explanation:

The execution in the option D is correct. This is because there is more than one reasonable criterion.

8 0
3 years ago
The process _______ boosted the production of fertilizers and strengthened the agriculture industry.
ANTONII [103]

Answer:

Process of Green Revolution

Explanation:

Green Revolution is a process which started its initiatives between the 1950s and 1960s, a technological research process, that increase agricultural productions through various systematic approach, which are different from the traditional methods. These technological approach includes:

1.  Utilization of new varieties with a high yield potential, in addition with adequate water  supply, pesticides and fertilizers.

2.  New methods of cultivation and mechanization

Hence, the process of Green Revolution boosted production of fertilizers and strengthen the agriculture industry as a whole.

6 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

3 0
3 years ago
For metallurgical reasons, it is desirable to melt the weld metal with minimum energy input. Which one of the following heat sou
stellarik [79]

Answer:

The correct option is;

(a) High power density

Explanation:

The power density of a material is the amount of power per unit volume of the material. Power density, in the context of transformers, fuel cells, batteries, motors, and power supply units, is measured with respect to the volume, and the units is given as W/m³

A system such as a capacitor with an high power density has the capacity to put out large energy amount from a small volume. A capacitor with a high power density, can produce the same power output as a car battery and is said to have a high power density.

8 0
3 years ago
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