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sertanlavr [38]
2 years ago
7

F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the

time it lands, then the relationship shown below can be used. Which of the four options below it correctly shows how this relationship can be expressed in terms of height, h?
Physics
1 answer:
stepladder [879]2 years ago
5 0

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

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2)Prepare a straight-line amortization table for the bonds' first two years.

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01/01/2019                                      61,950                           508,050  

06/30/2019                                      54,206                          515,794  

12/31/2019                                       46,462                         523,538  

06/30/2020                                       38,718                        531,282  

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                       Discount on bonds payable, Cr     (61950/8)  7743.75

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4) Record the interest payment and amortization on December 31:

Dec 31                 Bond interest expense, Dr               31969  

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                                    Cash, Cr                                24225

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