PHYSICS COLLEGE

Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31. They are issued at $508,050 when the market rate is 12%.1. Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31

Answers

Answer 1
Answer:

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:    

8 payments of $24,225:           $193,800    

Par value at maturity:                 $570,000    

Total repaid:                                   $763800 (193,800 + 570,000)  

Less amount borrowed:         $508050    

Total bond interest expense: $255750 (763800 - 508,050)

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

Semiannual Interest Period­ End; Unamortized Discount; Carrying Value

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  

12/31/2020                                         30,974                          539,026

3) Record the interest payment and amortization on June 30:

June 30            Bond interest expense, dr                         31969  

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

                                        Cash, Cr                     ( 570000*8.5%/2)  24225  

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

Dec 31                 Bond interest expense, Dr               31969  

                           Discount on bonds payable, Cr  7744  

                                    Cash, Cr                                24225

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A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.

Answers

Answer:

B = \mu_0((1)/(3) (J_0)/(r_0) r^2)

Explanation:

As the current density is given as

J = (J_0)/(r_0)r

now we have current inside wire given as

i = \int J(2\pi r)dr

i = \int (J_0)/(r_0) r(2\pi r)dr

i = 2\pi (J_0)/(r_0) \int r^2 dr

i = (2)/(3) \pi (J_0)/(r_0) r^3

Now by Ampere's law we will have

\int B. dl = \mu_0 i

B. (2\pi r) = \mu_0((2)/(3) \pi (J_0)/(r_0) r^3)

B = \mu_0((1)/(3) (J_0)/(r_0) r^2)

A periodic wave travels from one medium to another. Which pair of variables are likely to change in the process? A. velocity and wavelength B. velocity and frequency C. frequency and wavelength D. frequency and phase E. wavelength and phase

Answers

Answer:

A. velocity and wavelength

Explanation:

When a wave travels from one medium to another it undergoes refraction which results to the change in direction.

Refraction of a wave is one of the property of waves that occurs when a wave changes direction when it passes from one medium to another. This occurs as a result of bending of the wave which occurs since the mediums involved are of different density and refractive index.

Apart from the change in direction, refraction is accompanied by the change in wavelength of a wave and thus a change in speed.

Answer:

A. velocity and wavelength

Explanation:

Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.

Answers

Answer:

The lid becomes tighter

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The radius of a typical human eardrum is about 4.15 mm. Calculate the energy per second received by an eardrum when it listens to sound that is at the threshold of hearing, assumed to be 1.20E-12 W/m2

Answers

The energy per second received by an eardrum is 6.4884 * 10^(-17) watt

Calculation of the energy per second;

The area should be

= \pi r^2\n\n= 3.14 * 0.00415m\n\n= 5.407 * 10^(-5)m^2

Now

The power should be

= 1.2 * 10^(-12) * 5.407 * 10^(-5)\n\n= 6.4884 * 10^(-17) watt

Learn more about the energy here: brainly.com/question/14338287

Answer:

Power energy per second will be equal to 6.4884* 10^(-17)watt

Explanation:

We have given radius of human eardrum r = 4.15 mm = 0.00415 m

Intensity at threshold of hearing I=1.2* 10^(-12)w/m^2

Area is given by A=\pi r^2=3.14* 0.00415^2=5.407* 10^(-5)m^2

We know that power is given by P=I* A=1.2* 10^(-12)* 5.407* 10^(-5)=6.4884* 10^(-17)watt

So power energy per second will be equal to 6.4884* 10^(-17)watt

find the area of a right triangle with sides 15.0 cm, 20.0 cm, and 25.0 cm. express your answer with the correct number of significant figures

Answers

The area of this triangle can be calculated using herons formula since th three sides are given. It is expressed as:

A=sqrt( s(s-a)(s-b)(s-c))

where s is equal to a+b+c / 2

s=15 +20 +25 /2=30
A=sqrt( 30(30-15)(30-20)(30-25))
A= 150 cm^3

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_(initial) + PE_(initial) = KE_(final) + PE_(final)

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = (1)/(2)mv^(2) + 0

2gH = v^(2)

v = √(2* 9.8* 5) = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0* 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = (1)/(2)gt^(2)

t = \sqrt{(2h)/(g)}

t = \sqrt{(2* 2)/(9.8) = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56* 0.638 = 4.823\ m

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

mgh= (1)/(2)mv^2

v = √(2gh)

v = √(2* 9.8 * 5)

v = 9.89 m/s

now he perform elastic collision

v = (m_1v_1)/(m_1+m_2)

v = (65* 9.89)/(65+20)

v = 7.57 m/s

time taken by the skies to fall is

h = (1)/(2)gt^2

t = \sqrt{(2h)/(g)}

t = \sqrt{(2* 2)/(9.8)}

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m
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