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3 years ago

How many estimates are used in the calculation of depreciation?

2 answers:

8 0

Answer: The correct answer is : There are several methods to calculate depreciation, such as: straight line, activity or units produced, sum of annual digits and double share on the decreasing value.

Explanation: The most commonly used methods are, the linear method, where the original cost of an asset is divided by its estimated useful life to find the amount to be amortized each year and also the accelerated depreciation method is used in which more depreciation is provided the first years of useful life of the asset and for the remaining years a lower depreciation is assigned. This method in turn can be: doubly decreasing balance (200%), 150% decreasing balance, 125% decreasing balance and digits of the sum of the years.

xeze [42]3 years ago

3 0

Answer:

Two estimates

Explanation:

There are mainly two estimates used in the calculation of depreciation such as the useful life and the salvage value of an asset. The salvage value is defined as the predicted amount that will be obtained by a company from an asset when it is disposed at the end of the useful life of the particular asset. On the other hand, the useful life commonly refers to the estimation of how long the asset is useful for the company. This is different from the lifespan of the asset.

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A 1,000 kg car has 50,000 joules of kinetic energy what is its speed??

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3 years ago

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Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i

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<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

$m_{1}$ and $m_{2}$ will be now $2m_{1}$ and $2m_{2}$

$r$ will be now $2r$

Let's rewrite the equation (1) with this new values:

$F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}$ (2)

Solving and simplifying:

$F=4G\frac{m_{1}2m_{2}}{4r^2}$

$F=G\frac{m_{1}m_{2}}{r^2}$ (3)

As we can see, equation (3) is the same as equation (1).

So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>

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3 years ago

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At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a² + b²

r = 101.98 km, a = 20 km, b = 100 km

$2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} ) + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} ) + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 ) + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt} = \frac{1900}{101.98} = 18.63 \ km/h$

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3 years ago