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Anika [276]
3 years ago
9

On July 1, 2019, Friedman Inc. invested $720,000 in a mine estimated to have 900,000 tons of ore of uniform grade. During the la

st 6 months of 2019, 100,000 tons of ore were mined.
Required:


1) Calculate depletion cost per unit.


2) Prepare the journal to record depletion.
Business
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

A.$0.8

B.Dr Depletion account 80,000

Cr Coal mine Asset 80,000

Explanation:

Friedman Inc

A.

Depletion cost per unit = Amount invested / Estimated tons of ore

$720,000/900,000

=$0.8

B.

Journal entry

Dr Depletion account 80,000

Cr Coal mine Asset 80,000

Total depletion of ore = Number of ore ×Depletion cost per unit

=$0.8*100,000

=80,000

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Answer:

c. full employment

Explanation:

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So the answer is c. full employment

3 0
3 years ago
A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension withou
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Answer:

a. radius = 0.0006m = 0.6mm and length =0.393m = 393mm

b. frequency =377.86Hz

Explanation:

Given:

mass of steel= 4g = 0.004kg

density of steel = 7890kg/m3

tensile stress of steel 7.0x10⁸

tension load =900N

from the density, we will calculate for the Volume of the steel string

density = mass/volume

volume = mass/density = 0.004/7890 = 5.07 x 10⁻⁷ m³

from the tensile stress will can get the maximum base Area of the string ,

tensile stress = load/area =

7x10⁸ = 900/A

A = 900/7x10⁸ = 1.29x10⁻⁶ m²

Area = πr²

area/pi = 4.105x10-7

radius = 0.0006m = 0.6mm

volume = Area x length

length = vol/area =(5.07 x 10⁻⁷ m³)/1.29x10⁻⁶ m² = 0.393m

b. the highest possible frequency is given by:

F = \frac{\sqrt{\frac{T}{m/L} }}{2L}

where T= tension, m=mass, L=length

F = \frac{\sqrt{\frac{900}{0.004kg/0.393m} }}{2*0.393m}

= 297.04/0.786

frequency =377.86Hz

4 0
3 years ago
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All the 4 statements are correct.

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