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3 years ago

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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)

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Answer:

The condition is true when their voltage and current specifications with their impedance are matched or complementary to each other.

Explanation:

Solution

Yes it is possible or true to interface an IC with a different technology like the TTL to HCS12 ports. but the condition is that their current and voltage specifications should be matched and their impedance and power also should be matched.

What this implies is that both their voltage and current requirements should be complementary to each other so as their impedance.

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Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

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Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

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Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$

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$\Delta W = 2.043\times 10^{6}\,kJ$

The change in chemical energy from gasoline is:

$\Delta E = \frac{\Delta W}{\eta}$

$\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}$

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The changes in gasoline consumption is:

$\Delta m = \frac{\Delta E}{L_{c}}$

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'd' is seperation between thje plates

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For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

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