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Physics-

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Easy

Question

Asatellite in a force frees paces weeps stationary interplanetary dust at a rate $left parenthesis d M divided by d t right parenthesis equals alpha v$ The acceleration of satellite is

$\frac{- 2 α v^{2}}{M}$
$\frac{- α v^{2}}{M}$
$\frac{- α v^{2}}{2 M}$
$- α v^{2}$

The correct answer is: $fraction numerator negative alpha v to the power of 2 end exponent over denominator M end fraction$

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Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of oscillation of the system is

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Acceleration A and time period T of a body in S.H.M. is given by a curve shown below. Then corresponding graph, between kinetic energy (K.E.) and time t is correctly represented by

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