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A boy kicked a ball with initial speed of 19.6 m/s at 45° angle. Another boy who was standing 67.4 m away starts to run to meet the ball.what must be the speed of the boy in order to catch the ball before the ball hits the ground.

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state and prove benoulli"s theorem.give tow application of it?

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I just hate BERNOULLI
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The moon subtends an angle of 57' mins at the base line equal to the radius of the earth.What is the distance of the moon frm the earth.(radius of the earth = 6.4 x 10^{6 })
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D = 57' = 57' * 60
= 3240"
arc second to radian = 3240 * 4.85 * 10^{6 }
= 16587 * 10^{6 }
D= b/{tex}\theta{/tex}
D = 6.4 * 10^{6 }/ 16587 * 10^{6}
= 32/82935
= 3.85 * 10^{8 }
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Differentiate
d/dx(x^{3}.sin4x)

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Use chain rule for it,
{tex}{d\over dx}(x^3.sin4x){/tex}
{tex}= x^3.{d\over dx}(sin 4x) + sin4x. {d\over dx} (x^3){/tex}
{tex}= x^3.cos4x.{d\over dx}(4x) + sin4x.3x^2{/tex}
{tex}= 4x^3cos4x + 3x^2sin4x{/tex}
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how much below the surface of the earth does the acceleration due to gravity becomes 99% of its value at the earth's surface ? (radius of earth =6400km )

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Drive the expression for viscous force acting on spherical body of radius r moving with velocity v through viscous liquid of coefficient of viscocityη Using dimensional analysis.

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Discus the variation of acceleration due to gravity with altitude . How does the expression modify when h<<R?

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Ans. Let g be the value of acceleration due to gravity at the surface of earth and g' at a height h above the surface of earth. If the earth is considered as a sphere of homogeneous composition, then g at any point on the surface of the earth is given by:
{tex}g = {GM\over R^2}{/tex} …………(1)
Where M mass of earth, R is radius of Earth.
Now, we ll determine Value of g at the distance h from the surface of earth.
Then distance b/w center of earth and point = (R+h)
{tex}g' = {GM\over (R+h)^2}{/tex} ……………(2)
divide eq (2) by (1)
{tex}{g'\over g} ={ { GM\over (R+h)^2} \over {GM\over R^2}}{/tex}
{tex}=> {g'\over g}= {R^2\over (R+h)^2}{/tex}
{tex}=> {g'\over g}= {R^2\over R^2 (1+{h\over R})^2} {/tex}
{tex}=> {g'\over g}= {1\over (1+{h\over R})^2}{/tex}
{tex}=> {g'\over g}= {(1+{h\over R})^{2}}{/tex}
Expanding using Binomial Theorem and neglecting terms with higher power of R, we get
{tex}=> {g'\over g}=(12{h\over R}){/tex}
So, The value of acceleration due to gravity 'g' decreases with altitude 'h'.
When h<< R then expression is
{tex}=> {g'\over g}=(12h){/tex}
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Drive the expression for pressure of a gas in a container? using it relate kinetic energy with pressure?

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Two parallel force of magnitude 4 and 6 newton are acting on a body placed at 4m apart. Resultant force will be

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Convert 144km/ hr to m/s?

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Ans. {tex}144Km/h = {144\times 1000\over 3600} = 40m/s{/tex}
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Carbon14 decay with a half life of about 5,800 year. In a sample of bone, the ratio of carbon14 to carbon12 is found to be 1/4 of what it is in free air. This bone may belong to period about x centuries ago, where x is

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Why we take force are equal when we connect two springs in series ?

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You are given a cylinder once the cylinder rolled up to a inclined plane and the same cylinder is rolled down the same inclined plane.
Find direction of friction in both the cases

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Ans. The direction of the frictional force will be up the inclined plane in both the motions.
The friction acts opposite to the relative motion of the point of contact on ground. In both the cases the force acting on the point of cylinder in contact with the plane is {tex}mgSin\theta{/tex}.
As there is pure rolling the velocity and acceleration of the point of contact must be zero. So friction acts upwards along the inclined plane.
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Water drops are falling in regular interval of time from top of tower of ht. 9 m. If 4th drop starts to fall when 1st drop reaches at ground. Find position of 2nd and 3rd drop from top of tower

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Ans. Height of tower = 9m
Initial Velocity of drop u = 0
Acceleration Due to gravity = 10m/s
Let time taken by first drop to reach ground = t
Then, Using
{tex}S = ut +{1\over 2}at^2{/tex}
{tex}=> 9 = 0\times t + {1\over 2}(10)t^2{/tex}
{tex}=> 9 = 5t^2{/tex}
{tex}=> t^2 = {9\over 5} => t = {3\over \sqrt 5}{/tex}
Now, the fourth drop is just leaving the tap, the 2nd and 3rd drop are somewhere midway, and the first drop is hitting the ground. And the time interval for this situation to occur is {tex}{3\over \sqrt 5}{/tex}sec.
Since it says the drops are falling at regular intervals, therefore, the time must be equally divided in that time interval of {tex}{3\over \sqrt 5}{/tex} sec, giving {tex}{1\over \sqrt 5}{/tex} second of intervals between the drops.
So time taken by second drop is {tex} {2\over \sqrt 5} sec{/tex} and time taken by third drop is {tex}{1\over \sqrt 5}sec{/tex}.
Distance covered by second drop = {tex}{1\over 2} at^2 = {1\over 2}\times 10\times {4\over 5}{/tex}= 4 m
Distance covered by third drop = {tex}{1\over 2} at^2 = {1\over 2}\times 10\times {1\over 5}{/tex}= 1m
So, 2nd drop and 3rd drop are 4m and 1m away from top of tower respectively.
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A balloon start rising from the ground with an acceleration of 1.25m/s2 after 10s a stone is released from the balloon determine
Maximum height of stone from ground
Time taken by stone to reach ground

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Ans.
Initial Velocity of balloon u = 0
Acceleration a = 1.25m/s^{2}
Time t = 10s
Final velocity v = {tex}u + at = 0 + 1.25\times 10 = 12.5m/s{/tex}
Distance traveled in 10 s,
{tex}=> S = ut +{1\over 2}at^2{/tex}
{tex}=> S = 0\times 10 +{1\over 2}\times1.25\times 100{/tex}
=> S = 62.50m
Now, balloon will drop down with initial velocity, u'= v=12.5 m/s with acceleration, a'= g =10 m/s^{2} and cover distance 62.5 m(as it is moving downwards).
Using,
{tex}=> S = ut +{1\over 2}at^2{/tex}
{tex}=>  62.5 = 12.5t +{1\over 2}(10)t^2{/tex}
{tex}=>  62.5 = 12.5t 5t^2{/tex}
{tex}=> 5t^2  12.5t  62.5 = 0{/tex}
{tex}=> t^2  2.5t + 12.5 = 0{/tex}
On Solving, We get
t = 5.0 or 2.5
Time can not be negative so the required answer is 5.0 seconds. the stone will reach the ground in 5.0 seconds.
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How to prove work energy throem by calclues method

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Will the momentum remain constant if some external force acts on the system?

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Ans. If an external force is acted on a body to move it, the body accelerates. So the momentum cannot be a constant.
But, if the force cannot make a body to move like push of a boy on a wall, The body remains at rest and the momentum of the body remains zero.
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SI unit of viscosity

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The SI unit of viscosity is Poise.
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Variation of acceleration due to gravity due to rotation of earth

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The triple point of neon and carbon dioxide are24. 57k and 216.55k respectively. Express these temperature on the Celsius and Fahrenheit scales

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Please refer this image for your answer.
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Which point on the curve corresponds to elastic limit and yield point of the wire?

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3 moles of an ideal gas kept at constant temperature of 300 C are compressed from 4 litre to 1 litre. Calculate work done in this process.

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for isothermal process work done is nrtln(initial volume /final volume) .take r=8.314,t=300+273 k,n=3,vol. initial =4,vol final =1 and thencalculate to get answer 8603.6431 joules.
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Write the relation between two angles for which horizontal Ranges will be equal.

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Deduce the height at which the value of g is the same as at a depth of R÷2 ?

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What is magnetic & azimuthal quantum number.

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Ans. The Azimuthal Quantum number is a quantum number for an atomic orbital that determines its orbital angular momentum and describes the shape of the orbital. The azimuthal quantum number is the second of a set of quantum numbers which describe the unique quantum state of an electron (the others being the principal quantum number, following spectroscopic notation, the magnetic quantum number, and the spin quantum number). It is also known as the orbital angular momentum quantum number, orbital quantum number or second quantum number, and is symbolized as {tex}\varphi{/tex}
The Magnetic Quantum number, designated by the letter ml,[dubious ] is the third in a set of four quantum numbers (the principal quantum number, the azimuthal quantum number, the magnetic quantum number, and the spin quantum number) which describe the unique quantum state of an electron. The magnetic quantum number distinguishes the orbitals available within a subshell, and is used to calculate the azimuthal component of the orientation of orbital in space. Electrons in a particular subshell (such as s, p, d, or f) are defined by values of {tex}\varphi{/tex}(0, 1, 2, or 3). The value of m can range from {tex}\varphi \space to \space \varphi{/tex} inclusive of zero.
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A man in the lift standing on a weighing scale & the lift is moving upward. The velocity of lift is 10m/s & weight of man is 70 kg then what would be the reading on the scale.

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