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Let the equation of oscillation be given by  Velocity would be given as  Kinetic energy at an instant is given by Time Period  is given by The Average Kinetic Energy would be given as follows The potential energy at an instant T is given by  The Average Potential Energy would be given by We can see Kav = Uav
Apparent frequency striking the wall and getting reflected is The frequency emitted by the bats is  Speed of sound is v Speed of bat is 0.03v Frequency of sound as heard by the bat
Let us assume the earthquake occurs at a distance s. The origin of the earthquake is at a distance of 1960 km.
Frequency of SONAR =40 kHz Speed of enemy submarine vo=360 km h-1 = 100 m s-1 This is the frequency which would be observed and reflected by the enemy submarine but won't appear the same to the SONAR(source) as again there is relative motion between the source(enemy submarine) and the observer(SONAR) The frequency which would be received by the SONAR is
A=0.05 m Tension in the string is T=mg The speed of the wave in the string is v Angular frequency of the wave is Since at t=0, the left end (fork end) of the string x=0 has zero transverse displacement (y=0) and is moving along the positive y-direction, the initial phase is zero.  Taking the left to the right direction as positive we have Here t is in seconds and x and y are in metres.
(a) The pulse does not have a definite frequency or wavelength however the wave has definite speed given the medium is non-dispersive. (b) The frequency of the note produced by the whistle is not 0.05 Hz. It only implies the frequency of repetition of the pip of the whistle is 0.05 Hz,
The wavelength of the given wave is The points with the same displacements and velocity at the same instant of time are separated by distances . The points of the string which have the same transverse displacements and velocity as the  point at 5 s and 11 s would be at a distance of  from x = 1cm. Therefore all points at distances  from the point x=1cm would have the same transverse...
The displacement of oscillation of a point at x = 1 cm and t = 1 s is The general expression for the velocity of oscillation is k=0.005 cm-1 The velocity of propagation of the wave is The velocity of oscillation of point at x = 1 cm and t = 1 cm is not equal to the propagation of the wave.
At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring. Let the amplitude be A The angular frequency of a spring-mass system is always equal to  Therefore
A = 5 cm = 0.05 m T = 0.2 s At displacement x acceleration is  At displacement x velocity is  (a)At displacement 0 cm
A = 5 cm = 0.05 m T = 0.2 s At displacement x acceleration is  At displacement x velocity is  (a)At displacement 3 cm
A = 5 cm = 0.05 m T = 0.2 s At displacement x acceleration is  At displacement x velocity is  (a)At displacement 5 cm
Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is The period of Torsional oscillations would be
The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period. For damping factor b we have   x=x0/2 t=0.77s m=750 kg
Mass of automobile (m) = 3000 kg There are a total of four springs. Compression in each spring, x = 15 cm = 0.15 m Let the spring constant of each spring be k
Let the initial volume and pressure of the chamber be V and P. Let the ball be pressed by a distance x. This will change the volume by an amount  ax. Let the change in pressure be  Let the Bulk's modulus of air be K. This pressure variation would try to restore the position of the ball. Since force is restoring in nature displacement and acceleration due to the force would be in different...
Let one mole of a  substance of atomic radius r and density  have molar mass M Let us assume the atoms to be spherical Avogadro's number is  For Carbon For gold For Nitrogen For Lithium For Fluorine
Let the suspended particles be spherical and have radius r The gravitational force acting on the suspended particles would be The buoyant force acting on them would be  The net force acting on the particles become Replacing mg in equation (i) with the above equation we get The above is the equation to be derived
As per Graham's Law of diffusion if two gases of Molar Mass M1 and M2 diffuse with rates R1 and R2 respectively their diffusion rates are related by the following equation In the given question R1 = 28.7 cm3 s-1 R2 = 7.2 cm3 s-1  M1 = 2 g   The above Molar Mass is close to 32, therefore, the gas is Oxygen.
Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P1 = 1 atm = 76 cm of Mercury Let the crossectional area of the tube be x cm2 The initial volume of the air column, V1 = 15x cm3 Let's assume once the tube is held vertical y cm of Mercury flows out of it. The pressure of the air column after y cm of Mercury has flown out of the column P2 = 76 - (76 - y)...
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