Friday, June 26, 2009

Problem on Focal length

Topic:-Focal length

The focal length of an optical system is a measure of how strongly it converges (focuses) or diverges (defocuses) light. A system with a shorter focal length has greater optical power than one with a long focal length; that is, it bends the pencil of rays more strongly, bringing them to a focus in a shorter distance.

Question :-

A girl is applying make-up in a concave mirror. Her face is 35cm in front of the mirror. The image is 72cm behind the mirror. Using the mirror equation find the focal length of the mirror. What is the magnification of the image?

ANSWER:

Focal length of the mirror=68.108 cm, Magnification of the image=2.057

The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:


1/f=1/ do +1/ di

do=35 cm

di=-72 cm

1/f=1/35-1/72

Focal length = f=68.108 cm

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho).

m = - di/ do=-(-72)/35= 2.057

The magnification of the image=2.057.

Friday, June 19, 2009

Force,Types of force with an Example

Topic :- Force

In physics, a force is a push or pull that can cause an object with mass to change in velocity.

Types of Force :-


electromotive force
the force that, by reason of differences in potential, causes a flow of electricity from one place to another, giving rise to an electric current.
moment of force
the effect of a force exerted on a lever and about a fixed point.
reserve force
energy above that required for normal functioning. In the heart it is the power that will take care of the additional circulatory burden imposed by bodily exertion.
shearing force
a force exerted perpendicularly to a horizontal surface.

Here is a problem which shows,how to find force

1. 10 kg lawn mower is pushed along by a force applied at a 20° angle to the horizontal. If the mower accelerates at 3.4 m/s^2. what is the force?

ANSWER: 107.33N

Mass=10kg

Angle=20 °

Acceleration =3.4m/sec^2

F sin (theta) =mg

mg
F= ---------
sin θ

10*9.8
= ---------
sin (20)
98
= -------
sin (20)

F =107.33N (Answer)

I hope it helped you,for more physics help you can reply me.

Friday, June 5, 2009

Question on Frequencies of Sound Harmonics given Fundamental Frequency

Waves and Sounds is one of the basic concept in Physics which deals with wavelength, frequency and velocity of waves and sounds under uniform motion. Initial Frequency generated by any instrument is called as fundamental frequency and sound harmonics.

Topic : Frequencies of successisive Sound Harmonics

In the problem below, an instrument vibrates to produce sound harmonics and it is all about how to find out frequency of those sound harmonics.

Question : If a violin string vibrates at 440 Hz as its fundamental frequency, what are the frequencies of the first four harmonics?

Solution :

A violin is a both ends fixed, and so successive harmonics are simply multiples of the fundamental:
Therefore, f1 = 440Hz (fundamental frequency)
f2 = 2f1 = 2 x 440 = 880 Hz
f3 = 3f1 = 3 x 440 = 1320 Hz
f4 = 4f1 = 4 x 440 = 1760 Hz


For more explanations related to the concept can be found in physics help or science help.