General
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Physics-

A fighter plane enters inside the enemy territory, at time t equals 0 with velocity v subscript 0 end subscript equals 250 blank m s to the power of negative 1 end exponent and moves horizontally with constant acceleration a equals 20 m s to the power of negative 2 end exponent (see figure). An enemy tank at the border, spot the plane and fire shots at an angle theta equals 60 degree with the horizontal and with velocity u equals 600 blank m s to the power of negative 1 end exponent. At what altitude H of the plane it can be hit by the shot?

Physics-General

  1. 1500 square root of 3 m    
  2. 125 blank m    
  3. 1400 blank m    
  4. 2473 blank m    

    Answer:The correct answer is: 1500 square root of 3 mIf it is being hit then
    d equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals left parenthesis u cos invisible function application theta right parenthesis t
    or t equals fraction numerator u cos invisible function application theta minus v subscript 0 end subscript over denominator a divided by 2 end fraction

    therefore blank t equals fraction numerator 600 cross times fraction numerator 1 over denominator 2 end fraction minus 250 over denominator 10 end fraction equals 5 blank s
    H equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction cross times g t to the power of 2 end exponent
    equals 600 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 5 minus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 25
    H equals 2473 blank m

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    physics-General
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    A equals r subscript 1 end subscript plus c equals r subscript 1 end subscript plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator mu end fraction close parentheses
    rightwards double arrow r subscript 1 end subscript equals 75 minus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator mu end fraction close parentheses
    rightwards double arrow 75 minus 45 equals 3 0 to the power of o end exponent
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    rightwards double arrow r subscript 1 end subscript equals 75 minus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator mu end fraction close parentheses
    rightwards double arrow 75 minus 45 equals 3 0 to the power of o end exponent
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    physics-General
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    General
    physics-

    The two lines A and B shown in figure are the agraphs of the de Broglie wavelength l as function of fraction numerator 1 over denominator square root of V end fraction (V is the accelerating potential) for two particles having the same charge
    Which of the two represents the particle of heavier mass ?

    The two lines A and B shown in figure are the agraphs of the de Broglie wavelength l as function of fraction numerator 1 over denominator square root of V end fraction (V is the accelerating potential) for two particles having the same charge
    Which of the two represents the particle of heavier mass ?

    physics-General
    General
    physics-

    In a photoelectric experiment anode potential is plotted against plate current

    In a photoelectric experiment anode potential is plotted against plate current

    physics-General
    General
    physics-

    From an inclined plane two particles are projected with same speed at same angle theta, one up and other down the plane as shown in figure. Which of the following statements left parenthesis s right parenthesis is/are correct?

    Here, alpha equals 2 theta comma beta equals theta

    Time of flight of A is, T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application left parenthesis alpha minus beta right parenthesis over denominator g blank c o s blank beta end fraction
    equals fraction numerator 2 u sin invisible function application left parenthesis 2 theta minus theta right parenthesis over denominator g blank c o s blank theta end fraction equals fraction numerator 2 u over denominator g end fraction t a n invisible function application theta
    Time of flight of B is, T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    So, T subscript 1 end subscript equals T subscript 2 end subscript. The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two is zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is towards A B blank, therefore collision will take place between the two in mid air.

    From an inclined plane two particles are projected with same speed at same angle theta, one up and other down the plane as shown in figure. Which of the following statements left parenthesis s right parenthesis is/are correct?

    physics-General
    Here, alpha equals 2 theta comma beta equals theta

    Time of flight of A is, T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application left parenthesis alpha minus beta right parenthesis over denominator g blank c o s blank beta end fraction
    equals fraction numerator 2 u sin invisible function application left parenthesis 2 theta minus theta right parenthesis over denominator g blank c o s blank theta end fraction equals fraction numerator 2 u over denominator g end fraction t a n invisible function application theta
    Time of flight of B is, T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    So, T subscript 1 end subscript equals T subscript 2 end subscript. The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two is zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is towards A B blank, therefore collision will take place between the two in mid air.
    General
    physics-

    Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis

    Assuming particle 2 to be at rest, substituting in
    y equals x tan invisible function application theta minus fraction numerator g x to the power of 2 end exponent over denominator 2 u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction blank left parenthesis theta equals 0 degree right parenthesis

    We have negative h equals fraction numerator negative g over denominator 2 left parenthesis 4 v to the power of 2 end exponent right parenthesis end fraction
    or v equals square root of fraction numerator g over denominator 8 h end fraction end root
    Which is a straight line passing through origin with slopesquare root of fraction numerator g over denominator 8 h end fraction end root

    Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis

    physics-General
    Assuming particle 2 to be at rest, substituting in
    y equals x tan invisible function application theta minus fraction numerator g x to the power of 2 end exponent over denominator 2 u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction blank left parenthesis theta equals 0 degree right parenthesis

    We have negative h equals fraction numerator negative g over denominator 2 left parenthesis 4 v to the power of 2 end exponent right parenthesis end fraction
    or v equals square root of fraction numerator g over denominator 8 h end fraction end root
    Which is a straight line passing through origin with slopesquare root of fraction numerator g over denominator 8 h end fraction end root