Find the centre of gravity of the sector of a circle subtending an angle 2α at the centre of the circle? Ans: To find centre of gravity of the sector of a circle first draw the figure AOB is the sector of the circle subtending an angle 2α at the centre. Ox is the bisector of …

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]]>Ans: To find centre of gravity of the sector of a circle first draw the figure

AOB is the sector of the circle subtending an angle 2α at the centre. Ox is the bisector of angle AOB.

Equation of the curve is r=a, referred to O as pole and Ox as initial line, a being the radius of the circle.

Ox is the line of symmetry. Hence, the C.G. of the sector lies on Ox. Hence ȳ=0.

And x=2/3[( ʃ^{α}_{-α}r^{3}cosθdθ)/( ʃ^{α}_{-α}dθ)]

=2/3.a[( ʃ^{α}_{-α}cosθdθ)/( ʃ^{α}_{-α}dθ)]

=2a/3[( sinθ)^{α}_{-α}]/[(θ) ^{α}_{-α}]

= 2a/3[( sinα)/α]

So, r= a

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]]>Let us know what happens when football player kicks a ball. In this post we understand all component which affects the best play in a football match by Physics laws by an example. When player kicks a ball with an Example A football player kicks a ball at angle of 37⁰ to the horizontal with an …

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]]>Let us know what happens when football player kicks a ball. In this post we understand all component which affects the best play in a football match by Physics laws by an example.

A football player kicks a ball at angle of 37⁰ to the horizontal with an initial speed of 15 m/sec. Assuming that the ball travels in a vertical plane, calculate (a) the time at which ball reaches the highest point (b) the maximum height reaches (c) the horizontal range of the ball (d) the time for which the ball is in air.

**Solution:**

- Time taken by the ball to reach the highest point (u sin θ)/g = (15 x sin 37⁰)/9.8 = 0.92 sec.
- Maximum height reached

H max = (u^{2}sin^{2} θ)/2g = (15)^{2}(sin 37⁰)^{2}/2 x 9.8 = 4.16 m

- The horizontal range of the ball

R = u^{2} sin 2 θ /g = (15)^{2} sin (2 x 37⁰)/9.8 = 21.2 m

- Time for which the ball is in air

T = 2u sin θ /g = 2 x 15 sin 37⁰ /9.8 = 1.84 sec

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]]>What is logarithm? The term logarithm is basically derived from two words, “logas” and “arithmos”. Logas implies ratio while arithmos means number. The logarithm of any number to a given base is the power to which the base must be raised to obtain that number. Foe example, we know that 2 raised to power 5 …

]]>What is logarithm?

The term logarithm is basically derived from two words, “logas” and “arithmos”. Logas implies ratio while arithmos means number. The logarithm of any number to a given base is the power to which the base must be raised to obtain that number.

Foe example, we know that 2 raised to power 5 is equal to 32 i.e. 2^{5} =32. In the log form, this may be written as log 2 32 =5 i.e. logarithm of 32 to the base 2 is equal to 5. In general, a^{x} = N, then log_{a}N =x.

here is some basic logarithm.

In this type of logarithms, the base is e, where

e=1+(1/1!)+ (2/2!)+ (3/3!)+…… = 2.7182818

The natural logarithm is also abbreviated as/n.ie. we may write log_{e} N as /n N.

The common logarithm of a number is the power to which 10 must be raised in order to obtain that number i.e. in this type of logarithm, the base is 10.

**The natural logarithm may be changed to common logarithm using the following relation:**

Log_{e} N = 2.3026 log_{10} N or /n N = 2.3026 log_{10} N = 2.3026 log N.

Logarithm of a negative number is meaningless, because it does not exist.

Learn distance between two points.

Log_{a} mn = log_{a} m + log_{a} n

Proof: Suppose a^{x} =m and a^{y}=n Then log_{a} m =x and log_{a} n = y.

Hence a^{x} x a^{y} = mn or a^{x+y}= mn.

Or, log_{a} mn = x+y = log_{a}m + log_{a} n.

In general, we can also write that

log_{a} mnpq …. = log_{a}m+ log_{a}n+ log_{a}p+ log_{a}q+…

**Some fundamental formulas **

To find the coordinates of the point which divides in a given ratio (m1 : m2) the line joining two given points (x1, y1) and (x2, y2). Let P1 be the point (x1, y1) , P2 be the point (x2, y2) and P be the required point, so that we have P1P: PP2 :: m1 …

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]]>To find the coordinates of the point which divides in a given ratio (m_{1} : m_{2}) the line joining two given points (x_{1}, y_{1}) and (x_{2}, y_{2}).

Let P_{1} be the point (x_{1}, y_{1}) , P_{2} be the point (x_{2}, y_{2}) and P be the required point, so that we have

P_{1}P: PP_{2} :: m_{1} : m_{2}

Let P be the point (x, y) so that if P_{1}m_{1}, PM and P_{2}m_{2} be drawn parallel to the axis of y to meet the axis of x in m_{1} , M and m_{2}, we have Om_{1} =x_{1}, m_{1}P_{1} = y_{1}, OM = x, MP= y, Om_{2} = x_{2}, and m_{2}P_{2} = y_{2}

Draw P_{1}R_{1} and PR_{2}, Parallel to OX, to meet MP and m_{2}P_{2} in R_{1} and R_{2} respectively.

Then,

P_{1}R_{1} = m_{1}M =OM-Om_{1} = x-x_{1},

PR_{2}= Mm_{2}=Om_{2}-OM=x_{2}-x,

R_{1}P = MP – m_{1}P_{1}= y-y_{1},

R_{2}P_{2} = m_{2}P_{2} –MP =y_{2}y.

From the similar triangles P_{1}R_{1}P and PR_{2}P_{2}, we have

m_{1}/m_{2}=P_{1}P/PP_{2}=P_{1}R_{1}/PR_{2}=(x-x_{1})/(x_{2}-x)

m_{1}(x_{2}-x)=m_{2}(x-x_{1})

x= (m_{1}x_{2}+m_{2}x_{1})/(m_{1}+m_{2})

again, m_{1}/m_{2}=P_{1}P/PP_{2}=R_{1}P/R_{2}P_{2}=(y-y_{1})/9y_{2}-y’)

So that, m_{1}(y_{2}-y)=m_{2}(y-y_{1})

And hence, y=(m_{1}y_{2}+m_{2}y_{1})/(m_{1}+m_{2})

The coordinates of the point which divides P_{1}P_{2} internally in the given ratio m_{1}:m_{2} are therefore,

X = (m_{1}x_{2}+m_{2}x_{1})/(m_{1}+m_{2})

And Y = (m_{1}y_{2}+m_{2}y_{1})/(m_{1}+m_{2})

The proof of this statement is similar to that of the preceding article and is left as an exercise for the student.

*Note:The coordinates of the middle point of the line joining (x _{1}, y_{1}) to (x_{2}, y_{2}) are : (x_{1}+x_{2})/2 and y_{1} +y_{2})/2.*

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