Ask question

Ask question

All categories

3 years ago

5

Mobo, a wireless phone carrier, completed its first year of operations on October 31. All of the year's entries have been record

ed, except for the following: a. At year-end, employees earned wages of $6,000, which will be paid on the next payroll date, November 6. b. At year-end, the company had earned interest revenue of $3,000. It will be collected December 1 -4 Part 3 Show the accounting equation effects of each required adjustment. (Enter any decreases to Assets, Liabilities, or Stockholders' Equity with a minus sign.) ansaction Assets Liabilities Stockholders' Equity 6,000 Salaries and Wages Payable 4,900 6,000 alaries and Wages Expense nterest Receivable a. 4.,900 Acounts Payable b. terest Revenue

1 answer:

kenny6666 [7]3 years ago

3 0

Answer:

Explanation:

a) Salaries and wages expenses $6,000

To salary for wages payable $6,000

b) Interest Receivable a/c-d- $4,900

To interest revenue $4900

Assets = Liabiliteies +stockholder equity

a) 0 = 6000 + (-6000)

(No effect) (Increase) (Decrease)

0 =6000 -6000

b) 4900 = 0 + 4900

(no effect) (increase)

You might be interested in

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner’s 2.00 T field with his f

jok3333 [9.3K]

Answer:

Explanation:

Using Len's and Faraday formula

Faraday law states that a voltage is induced in a circuit whenever relative motion exists between a conductor and a magnetic field and that the magnitude of this voltage is proportional to the rate of change of the lux

Lenz law states that the direction of an induced emf is such that it will always opposes the change causing it

induce emf, e = -NΔФ/Δt

where Ф = BAcosФ and Ф = 0

N = number of turns, 1

B = 2 T

d = 2.20 cm = 0.022 m

radius = 0.022 m / 2 = 0.011 m

A area = πr² = 3.142 ( 0.011)² = 0.000380 m²

e = ( 0.00038 m²) ( 2 / 0.250 s) = 0.00304 V

b) Using ohm's law e = IR

since the the ring will be a metal e.g silver, and metals are good conductors of electricity, the resistances will be very low and the induce emf is low the temperature of the ring should not change significantly.

8 0

3 years ago

A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 °C. The air is first expanded isothermally to 500 kP

Semmy [17]

Answer:

Isothermal expansion W₁ =-37198.9 J

Polytropic Compression W₂ =-34872.82 J

Isobaric Compression W₃ = -6974.566 J

The net work for the cycle = -79046.29 J

Explanation:

Mass of air = 0.15 kg = 150 g

Molar mass = 28.9647 g/mol

Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air

PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³

For isothermal expansion we have

P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³

Therefore work done

W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J

Stage 2

Compression polytropically we have

$\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n$ where P₃ = 2 MPa

Therefore V₃ = $(\frac{1}{4} )^{\frac{1}{1.2} }*V_2$ = 1.6904×10⁻² m³

Work = W₂ = $\frac{P_2V_2-P_3V_3}{n-1}$ = -34872.82 J

$\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}$ or T₃ = $T_2*(\frac{P_3}{P_2})^\frac{n-1}{n}$ = 785.12 K

Isobaric compression we have thus

Work done W₃ = P(V₁ -V₃) = -6974.566 J

Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J

7 0

3 years ago

The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbac

bazaltina [42]

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

$\bold{Area =B \times D_f}$

$=6\times 5\\\\=30 \ ft^2$

In point b, Calculating the wetter perimeter.

$\bold{P_w =B+2\times D_f}$

$= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft$

In point c, Calculating the hydraulic radius:

$\bold{R=\frac{A}{P_w}}$

$=\frac{30}{16}\\\\= 1.875 \ ft$

In point d, Calculating the value of Reynolds's number.

$\bold{Re =\frac{4VR}{v}}$

$=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\$

$=750,000 V$

Calculating the velocity:

$V= \sqrt{\frac{8gRS}{f}}$

$= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\$

$\sqrt{f}=\frac{3.108}{V}\\\\$

calculating the Cole-brook-White value:

$\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\$

$\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})$

After calculating the value of V it will give:

$V= 25.18 \ \frac{ft}{s^2}\\$

In point a, Calculating the value of Froude:

$F= \frac{V}{\sqrt{gD}}$

$= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\$

$= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98$

The flow is supercritical because the amount of Froude is greater than 1.

Calculating the channel flow rate.

$Q= AV$

$=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\$

4 0

3 years ago

Define Plastic vs elastic deformation.

Snowcat [4.5K]

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

3 0

3 years ago

What is flow energy? Do fluids at rest possess any flow energy?

anzhelika [568]

Answer:

Flow energy is defined as, flow energy is the energy needed to push fluids into control volume and it is the amount of work done required to push the entire fluid. It is also known as flow work. Flow energy is not the fundamental quantities like potential and kinetic energy.

Fluid at state of rest do not possess any flow energy. It is mostly converted into internal energy as, rising in the fluid temperature.

8 0

3 years ago