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3 years ago

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Which of the following is a true statement?A. Meals, lodging, and incidental expenditures are only deductible if the taxpayer is

away from home overnight while traveling.B. Meals are deductible for an employee who is forced to work during the lunch hour.C. When a taxpayer travels solely for business purposes, only half of the costs of travel are deductible.D. If travel has both business and personal aspects, the cost of transportation is always deductible but the deductibility of lodging depends upon whether business is conducted that day.E. None of these is true.

1 answer:

aksik [14]3 years ago

5 0

Answer:

A.

Explanation:

Meals, lodging, and incidental expenditures are only deductible if the taxpayer is away from home overnight while traveling.

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Answer: Post conventional level

Explanation:

The Benjamin are at the post conventional level of the moral development as, the post conventional profound quality is the most worthy phase of profound quality in Kohl berg's model, where people have built up their very own arrangement of morals and ethics that they use to drive their conduct.

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Therefore, post conventional level is the correct option.

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3 years ago

Charles company used the percent of sales method to determine its bad debts expense. At the end of the current year, the company

fomenos

Answer:

$240,200

Explanation:

The computation of the account receivable amount reported in the balance sheet is given below:

Bad debt expense os

= $900000 × .50%

= $4,500

And,

Allowance for doubtful accounts after adjustment is

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= $4,800

Now

Balance of account receivable is

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= $240,200

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3 years ago

Cash dividends of $85,000 were declared during the year. Cash dividends payable were $10,000 and $15,000 at the beginning and en

Thepotemich [5.8K]

The amount of cash for the payment of dividends during the year is $80,000

What is the cash amount paid as dividends in the year?

The task at hand is to determine the amount of actual cash paid to shareholders as dividends in the year under review, which takes into consideration the amount of cash dividends declared in the year, the amount of dividends outstanding at the beginning of the year as well as the amount as at the end of the year.

$10,000 was unpaid at the beginning of the year, when declared during the year, which is $85,000 is added to it, we have $95,000 which the shareholders are expecting from the company.

The fact that at the end of the year, only $15,000 is unpaid means that out of the $95,000, $80,000 has been paid leaving us with an unpaid balance of just $15,000

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cash dividends paid=$10,000+$85,000-$15,000

cash dividends paid=$80,000

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2 years ago

The public relations nightmare from the U.S. Secret Service in 2015 has been attributed to a breakdown in which managerial funct

erica [24]

Answer: a. Controlling

Explanation:

The Controlling function in management is meant to ensure that employees in a company are acting in a manner that abides by the standards of the company or organization in question.

It works by managers ensuring that they check that employees are acting in the way they are to act and if they are not, corrective action must be meted out to stop the behavior.

The Secret Service had some embarrassing moments in 2015 with some agents being found drunk on a trip to Europe where they were assigned to President Obama's detail. Had supervisors been making sure that subordinates acted in a manner befitting of the secret service, the acts would have had a significantly less chance of happening.

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3 years ago

A point charge q1 = -9.6 μC is located at the center of a thick conducting spherical shell of inner radius a = 2.4 cm and outer

inessss [21]

Answer:

Part 1: The value of x component of electric field at point P is $-1.03 \times 10^7 \frac{ N}{ C}$

Part 2: The value of y component of electric field at point P is 0.

Part 3: The value of x component of electric field at point R is 0.

Part 4: The value of y component of electric field at point R is $-5.06 \times 10^8 \frac{ N}{ C}$ .

Part 5: The value of surface density at the outer edge of the shell is $-3.83 \times 10^{-4} C/m^2$ .

Part 6: None ,The field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of $r^2$ .

Part 7:The fields are equal as the charge on the outer shell does not affect the field on within the shell. ( $E_2=E_o$ )

Explanation:

Part 1

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$x= 8.4 cm = 0.084$ m is the location of P on x

So the value is given as

$E_x( P) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.084^2}\right) \\= -1.03 \times 10^7 \frac{ N}{ C}$

The value of x component of electric field at point P is $-1.03 \times 10^7 \frac{ N}{ C}$

Part 2

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$y = 0 cm = 0.0$ m is the location of P on y

So the value is given as

$E_y( P) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0$

The value of y component of electric field at point P is 0.

Part 3

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$x = 0 cm = 0.0$ m is the location of R on x

So the value is given as

$E_x( R) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0$

The value of x component of electric field at point R is 0.

Part 4

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$y =1.2 cm = 0.012$ m is the location of R on y

So the value is given as

$E_y( R) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.012}\right) \\=-5.06 \times 10^{8} N/C$

The value of y component of electric field at point R is $-5.06 \times 10^8 \frac{ N}{ C}$ .

Part 5

As

$\sigma = \frac{ q_{enclosed}}{ 4 \pi r^2}$ is the Surface charge density

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

So the value is given as

$\sigma_b = \frac{ q_{enclosed}}{ 4 \pi b^2}\\\sigma_b = \frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{ 4 \pi 0.041^2}\\\sigma_b = -3.83 \times 10^{-4} C/m^2$

The value of surface density at the outer edge of the shell is $-3.83 \times 10^{-4} C/m^2$ .

Part 6

<em>None because the field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of </em> $r^2$ .

Part 7

<em>The fields are equal as the charge on the outer shell does not affect the field on within the shell. (</em> $E_2=E_o$ )

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3 years ago