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zimovet [89]
3 years ago
12

Obama company has identified that bill clintonâs receivable account of $100 is uncollectible. What is the journal entry needed t

o write off the account under the allowance method?
Physics
1 answer:
beks73 [17]3 years ago
8 0

Answer:

The journal entry required to write off the account under the allowance method is Bad Debt Written Off.      

Explanation:

Accounts uncollectible are receivables, loans, or other debts that have virtually no chance of being paid. An account may become uncollectible for many reasons, including the debtor's bankruptcy, an inability to find the debtor, fraud on the part of the debtor, or lack of proper documentation to prove that debt exists.

So the next right steps would be to <em>Debit Allowance for Doubtful Debts </em>account and <em>Credit Accounts Receivable</em> for Bill Clinton.    

Cheers!

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A 4kg block and a 2kg block can move on horizontal frictionless surface. The blocks are accelerated by a +12-N force that pushes
Stolb23 [73]

Answer:

a) -4 N

b) +4 N

Explanation:

Draw a free body diagram for each block.

For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.

For the small block, there is 1 force, F pushing to the right.

There are also weight and normal forces in the vertical direction, but we can ignore those.

Sum of forces on the large block in the x direction:

∑F = ma

12 − F = 4a

Sum of forces on the small block in the x direction:

∑F = ma

F = 2a

2F = 4a

Substitute:

12 − F = 2F

12 = 3F

F = 4

The small block pushes on the large block 4 N to the left (-4 N).

The large block pushes on the small block 4 N to the right (+4 N).

4 0
4 years ago
A 12000 kg boat is moving 4.25 m/s. Its engine pushes 9200 N forward, but the current pushes back at 12,500 N. How much times do
Verizon [17]

Answer:

15.5 seconds

Explanation:

Apply Newton's second law:

∑F = ma

-12500 + 9200 = (12000) a

a = -0.275 m/s²

v = at + v₀

0 = (-0.275) t + 4.25

t = 15.5 s

It takes the boat 15.5 seconds to stop.

7 0
3 years ago
A spherical ball is dropped through a liquid, explain why it reaches terminal velocity.
Alekssandra [29.7K]

Probably because of the drag coefficient and the density of the liquid.

8 0
3 years ago
A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in
Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

\frac{GMm}{d^{2}}=\frac{mv^2}{d}

v=\sqrt{\frac{GM}{d}}

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}

v = 1.81 x 10^-4 m/s

7 0
4 years ago
On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

3 0
3 years ago
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