General
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Physics-

A point mass m is suspended from a light thread of length l comma fixed atO, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are

Physics-General

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    Answer:The correct answer is: T equals t e n s i o n comma blank W equals w e i g h t blank a n d blank F equals c e n t r i f u g a l blank f o r c e

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    A particle of mass m is rotating in a horizontal circle of radius R and is attached to a hanging mass M as shown in the figure. The speed of rotation required by the mass m keep M steady is

    To keep the mass M steady, let T is the tension in the string joining the two. Then for particle m comma
    T equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses i close parentheses
    For mass M comma
    T equals M g left parenthesis i i right parenthesis
    From Eqs. (i) and (ii)
    fraction numerator m v to the power of 2 end exponent over denominator R end fraction equals M g ⟹ v equals square root of fraction numerator M g R over denominator m end fraction end root

    A particle of mass m is rotating in a horizontal circle of radius R and is attached to a hanging mass M as shown in the figure. The speed of rotation required by the mass m keep M steady is

    physics-General
    To keep the mass M steady, let T is the tension in the string joining the two. Then for particle m comma
    T equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses i close parentheses
    For mass M comma
    T equals M g left parenthesis i i right parenthesis
    From Eqs. (i) and (ii)
    fraction numerator m v to the power of 2 end exponent over denominator R end fraction equals M g ⟹ v equals square root of fraction numerator M g R over denominator m end fraction end root
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    A particle is projected from a point A with velocity u square root of 2 at an angle of 45 degree with horizontal as shown in figure. It strikes the plane B C at right angles. The velocity of the particle at the time of collision is

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    Then, u square root of 2 cos invisible function application 45 degree equals v sin invisible function application 60 degree
    open parentheses u square root of 2 close parentheses open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses equals fraction numerator square root of 3 v over denominator 2 end fraction blank therefore v equals fraction numerator 2 over denominator square root of 3 end fraction u

    A particle is projected from a point A with velocity u square root of 2 at an angle of 45 degree with horizontal as shown in figure. It strikes the plane B C at right angles. The velocity of the particle at the time of collision is

    physics-General
    Let v be the velocity at the time of collision

    Then, u square root of 2 cos invisible function application 45 degree equals v sin invisible function application 60 degree
    open parentheses u square root of 2 close parentheses open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses equals fraction numerator square root of 3 v over denominator 2 end fraction blank therefore v equals fraction numerator 2 over denominator square root of 3 end fraction u
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    Assertion: At constanpressurfothe change H2O(s)→H2O(gwordoninegative.
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    Ivieof the signof DG° or the following reactions:
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    Thevalueofenthalpychange(DH)forthe reactionC2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l) at 27°Cis–1366.5kJmol–1.The valueofinter- nalenergychangefortheabovereactionat this temperaturewillbe-

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    Fromthethermochemicalreactions,C(graphite)+½O2 →CO;DH=–110.5KJ CO+1/2O2 →CO2;DH=83.2KJ Theheatofreactionof C(graphite)+O2 →CO2is-

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    chemistry-General
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    chemistry-

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