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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.

Posted by raghav aggarwal (Apr 27, 2017 1:20 p.m.) (Question ID: 5043)

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

Posted by siva selvam963 (Apr 24, 2017 4:19 p.m.) (Question ID: 5024)

  • Answers:
  • I just hate BERNOULLI

    Answered by Aashka Shah (Apr 24, 2017 5:03 p.m.)
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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  )

pls answer fast   

Posted by Manav Dhora (Apr 21, 2017 12:57 p.m.) (Question ID: 4979)

  • pls reply fast

     

    Posted by Manav Dhora (Apr 21, 2017 1:04 p.m.)
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  • Answers:
  • 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 * 108

    Answered by Manav Dhora (Apr 21, 2017 3:54 p.m.)
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Differentiate

d/dx(x3.sin4x) 

Posted by M G (Apr 14, 2017 2:37 p.m.) (Question ID: 4880)

  • Answers:
  • 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}

    Answered by Payal Singh (Apr 17, 2017 6:47 a.m.)
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how spectograph work?

 

Posted by naveen vishwakarma (Apr 07, 2017 3:34 p.m.) (Question ID: 4765)

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 )

Posted by Prerna Pihu (Mar 27, 2017 5:02 p.m.) (Question ID: 4552)

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.

 

Posted by Innocent Geeta (Mar 24, 2017 11:19 p.m.) (Question ID: 4492)

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

Posted by Innocent Geeta (Mar 24, 2017 11:03 p.m.) (Question ID: 4490)

  • Answers:
  • 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}=(1-2{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}=(1-2h){/tex}

     

    Answered by Naveen Sharma (Mar 26, 2017 10:32 a.m.)
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Drive the expression for pressure of a gas in a container? using it relate kinetic energy with pressure?

Posted by Innocent Geeta (Mar 24, 2017 11 p.m.) (Question ID: 4489)

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

Posted by Deepak James (Mar 23, 2017 10:48 a.m.) (Question ID: 4435)

Convert 144km/ hr to m/s?

Posted by Devidas gaude (Mar 20, 2017 8:05 p.m.) (Question ID: 4239)

  • Answers:
  • Ans. {tex}144Km/h = {144\times 1000\over 3600} = 40m/s{/tex}

    Answered by Naveen Sharma (Mar 20, 2017 8:28 p.m.)
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Define gravition.

Posted by Vishal Kumar (Mar 19, 2017 9:20 a.m.) (Question ID: 4089)

Carbon-14 decay with a half life of about 5,800 year. In a sample of bone, the ratio of carbon-14 to carbon-12 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

Posted by ATHARVA KALE (Mar 18, 2017 5:05 p.m.) (Question ID: 4063)

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

Posted by Vivek Kumar (Mar 17, 2017 1:08 p.m.) (Question ID: 3985)

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

Posted by Vivek Kumar (Mar 17, 2017 11:19 a.m.) (Question ID: 3972)

  • Answers:
  • 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.

    Answered by Naveen Sharma (Mar 17, 2017 12:37 p.m.)
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state and prove bernoullis threom?

Posted by prakhar tripathi (Mar 17, 2017 7:25 a.m.) (Question ID: 3960)

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

Posted by Dr-Sachin Khatri (Mar 16, 2017 8:15 p.m.) (Question ID: 3940)

  • Sir please ans quickly 

    Posted by Dr-Sachin Khatri (Mar 16, 2017 8:45 p.m.)
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  • Answers:
  • 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.

    Answered by Naveen Sharma (Mar 17, 2017 9:40 a.m.)
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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

Posted by Dr-Sachin Khatri (Mar 16, 2017 3:55 p.m.) (Question ID: 3921)

  • Answers:
  • Ans. 

    Initial Velocity of balloon u = 0 

    Acceleration a = 1.25m/s2

    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/s2 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.

     

    Answered by Naveen Sharma (Mar 16, 2017 7:35 p.m.)
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How to prove work energy throem by calclues method

 

Posted by shailesh mohan (Mar 13, 2017 10:32 p.m.) (Question ID: 3707)

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

Posted by priyadarshan sundar (Mar 11, 2017 7:10 p.m.) (Question ID: 3531)

  • Answers:
  • 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.

    Answered by Naveen Sharma (Mar 16, 2017 9:47 a.m.)
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SI unit of viscosity

Posted by Durgesh Verma (Mar 10, 2017 7:26 p.m.) (Question ID: 3464)

  • Answers:
  • The SI unit of viscosity is Poise.

    Answered by Shweta Gulati (Mar 12, 2017 7:10 p.m.)
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Variation of acceleration due to gravity due to rotation of earth

Posted by Aryaman Shrivastava (Mar 10, 2017 4:13 a.m.) (Question ID: 3406)

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

Posted by Pradyumn Kumar (Mar 07, 2017 9:57 a.m.) (Question ID: 3226)

  • Answers:
  • Please refer this image for your answer.

    Answered by Shweta Gulati (Mar 09, 2017 11:27 a.m.)
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Which point on the curve corresponds to elastic limit and yield point of the wire?

Posted by Purav Jha (Mar 06, 2017 3:46 p.m.) (Question ID: 3172)

Factor  that rigid body depend

Posted by Kartikey Singh Chauhan (Mar 05, 2017 10:16 p.m.) (Question ID: 3144)

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. 

Posted by Muskan Goel (Mar 05, 2017 8:12 p.m.) (Question ID: 3132)

  • Answers:
  • 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.

     

    Answered by sambeet kar (Mar 05, 2017 8:39 p.m.)
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Write the relation between two angles for which horizontal Ranges will be equal.

Posted by sparsh anand (Mar 05, 2017 10:32 a.m.) (Question ID: 3088)

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

Posted by Anshpreet Nagi (Mar 04, 2017 9:17 p.m.) (Question ID: 3065)

What is magnetic & azimuthal quantum number.

Posted by Rajat Khurana (Mar 04, 2017 5:50 p.m.) (Question ID: 3052)

  • Answers:
  • 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.

    Answered by Naveen Sharma (Mar 04, 2017 7:01 p.m.)
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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.

Posted by Rajat Khurana (Mar 04, 2017 5:48 p.m.) (Question ID: 3050)

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