Semester Final Examination

Name : Melvika Anggarini
NIM   : RSA1C311010
Prodi  : Pend. Fisika PGSBI

1.      Design a simple research about the kinetic energy associated with your daily life and arrange your report in accord with the sixth science process skills you have already understood

       Yesterday, i saw my nephew threw a small rock toward his friend in front of him.  And i observed the moving of stone.  When the stone release from my nephew hand, the stone hit  the forehade of my nephew’s friend. Than I think,  to hit the forehead of my nephew’s friend, a stone need some energy. So, when  the stone threw from my nephew hands, it have a velovity to move and have energy to hit. Energy that the stone have because of its velocity called kinetic energy.

       Energy is capacity system to do work.  There are many kind of energy in this world, one of them is kinetic enrgy. Kinetic energy is the energy  in moving object or the energy that happend because of velocity an object. If we see an object with velocity v and mass m, so we can say that the kinetic energy is equals with half times the mass of the object times square the speed of the object. In symbols :

E_K = (1/2)mv²

Where :      Ek = Kinetic Energy ( Joule or J )

             m = Mass of object ( Kilogram or Kg )

        v = velocity of an object ( meter/ sekon or m/s )

So the kinetic energy is proportional to the mass of the object m and proportional to the square of the velocity. If the masses become twice the kinetic energy increase twice as well, and if speed is increased to twice the kinetic energy will increase to four times. So if we back to the problem above, if we increase the mass of stone twice, the kinetic energy will increase twice and it’s mean the stone will hurter than initial mass of stone. In other hand, if we inrease the velocity of object twice, the kinetic energy will raise twice too, so same like above,  it;s mean the stone will hurt the forehead my nephew friends more than intial velocity of stone.

Example :

How much kinetic energy does an object have if its mass is 5.0 kg and it is moving at a speed of 4.0 m/s?

Solution

Known : m = 5 kg

               V =  4 m/s

Ask        : Ek ?

Answer  : Ek = (1/2)mv²

                      = ½ x 5 kg x ( 4 m/s) ²

    Ek = 40 Joule

So, how about kinetic energy in deceelerate object?. I think a deceelerate object sill have kinetic energy because the object still have velocity but decrease. And because velocity is decreased the kinetic energy will decreased to until touching zero if velocity is zero.

2.      Describe an experiment which associated with the following terms : pendulum, frictional force and  highest point, then enter the terms into the procedures so that these terms included in your experiment.

I believe every one ever heard about constant acceleration of gravity. Constant acceleration of gravity in earth, moon, and other planet is different each other . In earth, we already know the value  constant acceleration of gravity is 9,8 m/s square or usually high school book write it is 10 m/s square. To proove the value of constant acceleration of gravity is 9,8 m/s square, we can do a simple experiment using a pendulum. In this experiment, In analyzing the simple pendulum motion, the friction force air we ignore because it will so complicated if we calculate friction force, and mass of  the rope is very negligible so we can ignore it relative to the ball. In do experiment with pendulum in highest point 15 degrees and some length of rope to take 10 swings, and calculate the result with this equation we can get constant acceleration of gravity : g = 4 phi^2 l / T

Now we will study about the experiment, check it out J

What we need?

1. Ball mattress 1 meter

2. stopwatch

3. gravel

4. arc

5. A4 paper

6. stationery

What is the procedure?

1.      Tying stone with rope

2.      Swinging a stone that has been tied to the 20-cm

3.      Observing the pendulum swings until have harmonic motion  in the  highest point is  15 degrees

4.      Counting up to 10 vibrations a stopwatch

5.      Take note of the time required for 10 vibrations

6.      Repeating the experiment from 2 to 5 with a length of rope 40 cm and 60 cm and 80 cm.

7.      Noting the results of experiments on the observation table.

8.      Calculate the constant acceleration of gravity using equation given before.

9.      Compare the value conctant acceleration of gravity known with result of experiment.

Result

The value of constant acceleration of gravity that we get from experiment is little different with the wellknown value. It may be happend because maybe there are some error in experiment, human error or instrumental error. But the result is neaar 9 – 10 m/s square.

3.      Make a simple calculation about electrical using the formula your well known. Write down your solution, and explain it by using your sentence ( change the number or symbol into your own word )

Current and charge

            We can see that when charged particles flow past a point in a circuit, we say that there is a current in the circuit. The charge flow past the point need time.  Electrical current is measured in ampere ( A ), Charged measured in Coulumb ( C ) and time measured in sekon ( s ).  So, we can get relationship between charge, current and time written as following word equation  :

            Charge = current x time

The formula written in symbol :

            delta Q = I. deltat

Where : delta Q = The amount of charge  which flows during a time interval  ( C )

              I           = Current ( A )

              delta t   = change or interval of time ( s )

From the equation above we can say that :

“One coulumb is the amount of charge which flows past a point in a circuit a time of 1 s when the current is 1 A.”

“The Amount of charge is proportional with electrical current and invers with interval of time, in higher value of electrical current and lower interval of time will produce bigger amount of charge”.

Example :

1.      There is a current of 10 A through a lamp for 1 hour. Calculate how much chrarge flows through the lamp in this time.

Solution

Known  :  I         =  10 A

               delta t  =  1 hour  = 3.600 s

Ask         : delta Q ?

Answer :  delta Q = I x  delta t 

                            = 10 x 3.600 s

               delta Q  = 36.000 C

So, in this problem the amount of charge which flows past in a circuit with current is 10 A in 1 hour is 36.000 C

2.      Calculate the current in a circuit when a charge of 180 C passes a point in a circuit in 2 minutes.

Solution

Known   :  delta Q  = 180 C

                 delta t    = 2 minutes = 2 x 60 s = 120 s

Ask         : I ?

Answer :    I = delta Q  / delta t     

                  I =  180 C / 120 s

                  I = 1,5 A

So, in this problem to make 180 C charge passes a point in 2 minutes, we need 1,5 A electric current

4. Make an essay about sound as a wave. Then show at least three evidence and examplifying about it.

Sound is one kind of wave, a longitudinal wave exactly.  Longitudinal waves are waves in which the motion of the individual particles of the medium is in a direction that is parallel to the direction of energy transport. Example of longitudinal waves are wave created in slinky and sound wave in the air. In sound wave propagation, there are “rapatan” and “ renggangan”, that made by intermediet partisles waves. When the sound waves propagating in air, intermediaries are particles in the air. Sound waves can not propagate in a vacuum because there is no air particles in the air.

Sound as  waves have properties similar to the properties of wave, such as :

a. Can be reflected (reflection)

Sound can reflect  occuring  when sound meet hard surfaces, such as stone walls  surfaces, cement, steel, glass and zinc.

example:

- Our voices will louder in the cave due to the reflection of sound on the wall of the cave.

- Our voices inside the building or music studio that does not use a silencere will louder   than in the outside

      Location of the walls of the room were too close causing reflected sound is not enough time to propagate, so that the original sound  and reflected sound heard simultaneously. This proves the reflected sound reinforces the original sound. That's why the voice in music studios louder  than outside. This explanation also same with  the situation voice in the cuve.

b. Can be refracted (refiaksi)

Refiaksi is bent  direction of path wave after passing the boundary between two different medium.

Example:  

- At night the sound of thunder sounded louder than during the day because of the   refraction of sound waves.

        During the day, the air in the upper layers colder than the bottom layer. The propagation of sound in cold temperatures is smaller than the heat. Thus, the speed of sound in the upper air layer is smaller than the speed of sound in air layer below, because the medium more dense on the top layer of the medium on the bottom layer. So, during the day, the sound of thunder travels from the upper air layers toward lower air layer would refracted away from the normal line (Figure 3.2a).

   

Figure 3.2. Refraction of sound waves

At night, the opposite happens, the air in the lower layers (near the ground) is cooler than the air on the top layer. Thus, the speed of sound in the lower layer is smaller than the upper layer, because the medium on the top layer is less dense than the medium on the bottom layer. So, at night, the sound of thunder travels from the upper air layer leading to lower air layer (denser medium) will be refracted normal line approach (Figure 3.2b). Refraction sound of thunder approaching normal line at night is what causes the sound of thunder nearer your home, and as a result you hear the thunder louder.

c. Can be combined (interference)

Same as light interference, sound interference also requires two coherent sources of sound.

Example: Two loudspeakers are connected to a signal generator (audio frequency generators) can serve as two coherent sources of sound.

          If each loudspeaker produce sound that have same fase and same deviation and placed in a distance. So the combined of two sound will produce maximum interference  Interfernce maximum can reinforce sound, so in this case, the combined two sound can produce louder sound to be source sound.

d. Can be bent (diffracted)

Diffraction is flexing sound waves when passing through a narrow slit.

Example: We can hear the sound of the room of different and closed, because the sound through narrow slits that are passable sound.

       As we se above, diffraction is flxing sound when passing narrow slit, so in this case, in the different and closed room, we still can hear voice from another room because sound can flexing through narrow slit like ventilation in the room. So because this flexing sound or diffraction process the voice from another room still we hear.