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3 years ago

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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

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:

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The difference in quantity between the add and full marks on an engine oil dipstick is typically

svetoff [14.1K]

Answer:

1 quart (0.9 liters).

Explanation:

A proper inspection of various systems and components in a vehicle at regular intervals is very important and necessary because it helps to ensure that the vehicle is in a safe and reliable condition.

Generally, these inspection includes tyres, lighting systems, fan belts, shock absorbers, fluid (oil and water) level, etc. If any fault or concern is detected in the course of an inspection, it should be noted for quick repair or servicing by an expert technician.

All automobile engine requires an adequate amount of engine oil as a lubricant so as to mitigate friction and enhance proper functionality of the vehicle. Thus, the proper functionality of an engine is largely dependent on the level of the engine oil; it shouldn't be too low or high.

Basically, the engine oil should be checked at regular intervals (periodically) and should be on the level indicated or chosen by the manufacturer of the vehicle.

A dipstick is designed to be used for checking the engine oil level in a vehicle and it is marked with lines indicating minimum and maximum, low and high or add and full.

The difference in quantity between the add and full marks on an engine oil dipstick is typically 1 quart (0.9 liters).

5 0

3 years ago

A 16-lb solid square wooden panel is suspended from a pin support at A and is initially at rest. A 4-lb metal sphere is shot at

lbvjy [14]

Answer:

velocity = 0.6 ft/s

Explanation:

8 0

3 years ago

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c

Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice $C_p=2.03\ KJ/kg.K$

Latent heat for ice H=336 KJ/kg

Specific heat for ice $C_p=4.187\ KJ/kg.K$

We know that sensible heat given as

$Q=mC_p\Delta T$

Heat for -15F to 32 F:

$Q=mC_p\Delta T$

$Q=9\times 2.03(32+15) KJ$

Q=858.69 KJ

Heat for 32 Fto 200 F:

$Q=mC_p\Delta T$

$Q=9\times 4.187(200-32) KJ$

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So 7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0

3 years ago

A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 = (16/3) * 10^2 V

c) E = 64/15 J

d) dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(c) the ﬁnal energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

Q_initial = 1.5*Q_1

Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

V_2 = V_1 = Q_1 / C_1

= 10.67*10^-3 / 20*10^-6

= (16/3) * 10^2 V

c)

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

d)

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

5 0

4 years ago

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

Inessa05 [86]

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

$\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}$

Now we can use the equation that defines the percentage of increase, in this case for speed

$i=(\frac{V2}{V1} -1) 100$

Now we use the equation obtained in the previous step, and replace values

$i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65$

the percent increase in the velocity of air is 25.65%

6 0

3 years ago