A kind of foliated metamorphic rock a. coal b. limestone c. marble d. slate

At the beginning of 2016, robotics inc. acquired a manufacturing facility for $13.1 million. $10.1 million of the purchase price

ki77a [65]

$ 376153 is depreciation on the building for 2018.

<u>Explanation:</u>

Given data:

Original cost = $13.1 million

Residual value = $2.1 million

Useful life = 20 -year

Depreciation on the building for 2016 = $=\frac{\text {original cost-residual value}}{\text { Useful life }}$

By substituting the given values, we get,

$=\frac{(\$ 13.1 \text { million }-\$ 2.1\text { million) } }{20 \text { years }}=\frac{11.1 \text { million }}{20 \text { years }}$ = 5500000

Depreciation for 2016 $5500000

Depreciation for 2017 $5500000

Depreciation at Dec 31, 2017 $ 5500000

In 2018, the estimates of useful life and residual value were changed to 15 years and $610000, but 2 yrs have already passed, so there're 13 more yrs to go.

Depreciation on the building for 2018,

$=\frac{\text { book value at the end of } 2016-\text {residual value}}{\text { useful life }}$

= $=\frac{\$ 5500000-\$ 610000}{13}=\frac{4890000}{13}=\$ 376153$

4 0

3 years ago

Explain why an object’s weight is dependent upon where in the universe it is located.

strojnjashka [21]

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the law of universal gravity

$F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]$

Now replacing the values in the equation we have:

$F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]$

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

$g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]$

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a radius of 71492 [km]. Find the force exerted by the planet upon the person.

$F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} } \\F= 1734.73[N]$

We will calculate the value of the gravity accelaration of Jupiter. Let's check

$g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\$

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

6 0

3 years ago

How can an electron in an atom lose energy to go from a higher energy level to a lower energy level? select one:

Romashka-Z-Leto [24]

I belive that the answer is b.

8 0

3 years ago

Please help me there

frozen [14]

Answer:

when switch is off no electricity will flow and then the circuit is called an open circuit

Explanation: