20 examples of scalar quantity​

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

Answer:

It's impossible for an ideal heat engine to have non-zero power.

Explanation:

Option A is incomplete and so it's possible.

Option B is possible

Option D is related to the first lae and has nothing to do with the second law.

Hence, the correct option is C.

The ideal engine follows a reversible cycle albeit an infinitely slow one. If the work is being done at this infinitely slow rate, the power of such an engine is zero.

We can also stat the second law of thermodynamics in this manner;

It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat energy from a colder object to a hotter one.

This statement is known as second form or Clausius statement of the second law.

Thus, it is possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work input.

Final answer:

According to the second law of thermodynamics, it is impossible for heat energy to flow from a colder body to a hotter body, for an ideal heat engine to have an efficiency of 99%, and for a physical process to yield more energy than what is put in.

Explanation:

According to the second law of thermodynamics, it is impossible for heat energy to flow from a colder body to a hotter body. This is because heat naturally flows from a region of higher temperature to a region of lower temperature. This principle is what allows us to effectively use heat for various purposes, such as in heat engines.

An ideal heat engine is a theoretical construct used to study the efficiency of engines. The second law of thermodynamics states that no heat engine can have an efficiency of 100%, so it is impossible for an ideal heat engine to have an efficiency of 99%. This is due to the losses in heat transfer and other thermodynamic processes.

The second law of thermodynamics also implies that in any physical process, the total energy cannot increase. It is impossible for a physical process to yield more energy than what is put in. This principle is central to understanding energy conservation and the limitations of energy conversion.

Learn more about Second Law of Thermodynamics here:

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Answer:

The spring constant k is

Solution:

As per the question:

Length of the solid cylinder, L = 500 mm =

Diameter pf the cylinder, D = 2 cm = 0.02 m

As the radius is half the diameter,

Radius, R = 1 cm = 0.01 m

Young's Modulus, E = 17.4 GPa =

Now,

The relation between spring constant, k and Young's modulus:

where

A = Area

Area of solid cylinder, A =

k =

Young's modulus, E is the ratio of stress and strain

And

Stress =

Strain =

Also

Force on a spring is - kL

Therefore, we utilized these relations in calculating the spring constant.

Only ' A ' is always true. (choice-1)

' B ' is not true when you drop a red hot spoon into cold soup.

' C ' is not true when you drop a red hot marble into a cool swimming pool.

Answer:

The answer is 1.42 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

Force = 0.145 × 9.81 = 1.42245

We have the final answer as

1.42 N

Hope this helps you