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Maths-

The value of k for which the lines fraction numerator x plus 1 over denominator negative 3 end fraction equals fraction numerator y plus 5 over denominator 2 k end fraction equals fraction numerator z minus 4 over denominator 2 end fraction and fraction numerator x minus 3 over denominator 3 k end fraction equals fraction numerator y minus 2 over denominator 1 end fraction equals fraction numerator z plus 1 over denominator 7 end fraction are perpendicular is

Maths-General

  1. 3    
  2. -1    
  3. 2    
  4. -2    

    Answer:The correct answer is: 2

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    A boy begins to walk eastward along a street infront of his house and the graph of his displacement from home is shown in the following figure. His average speed for in the whole-time interval is equal to

    D i s p l a c e m e n t blank f o r m blank 0 blank t o blank 5 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 5 blank t o blank 10 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 0 blank t o blank 15 blank s blank equals blank minus blank 20 blank m
    A n d blank d i s p l a c e m e n t blank f r o m blank 15 blank t o blank 20 blank s equals 0 blank m
    therefore N e t blank d i s p l a c e m e n t equals 40 plus 40 minus 20 plus 0 equals 60 blank m
    T o t a l blank t i m e blank t a k e n equals 5 plus 5 plus 15 plus 5 equals 30 blank m i n.
    H e n c e comma blank a v e r a g e blank s p e e d equals fraction numerator d i s p l a c e m e n t blank open parentheses m close parentheses over denominator t i m e blank open parentheses m i n close parentheses end fraction equals fraction numerator 60 over denominator 30 end fraction equals 2 m blank m i n to the power of negative 1 end exponent.

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    D i s p l a c e m e n t blank f o r m blank 0 blank t o blank 5 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 5 blank t o blank 10 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 0 blank t o blank 15 blank s blank equals blank minus blank 20 blank m
    A n d blank d i s p l a c e m e n t blank f r o m blank 15 blank t o blank 20 blank s equals 0 blank m
    therefore N e t blank d i s p l a c e m e n t equals 40 plus 40 minus 20 plus 0 equals 60 blank m
    T o t a l blank t i m e blank t a k e n equals 5 plus 5 plus 15 plus 5 equals 30 blank m i n.
    H e n c e comma blank a v e r a g e blank s p e e d equals fraction numerator d i s p l a c e m e n t blank open parentheses m close parentheses over denominator t i m e blank open parentheses m i n close parentheses end fraction equals fraction numerator 60 over denominator 30 end fraction equals 2 m blank m i n to the power of negative 1 end exponent.
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    The displacement-time graph of a moving object is shown in figure. Which of the velocity-time graphs shown in figure could represent the motion of the same body?

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    Think of the lope of the given displacement-time graph at different points and you would arrive at the correct answer.
    Alternatively, look at the self-illustrative figure.
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    The number of solutions of the equation Sin space 3 alpha equals 4 Sin space alpha times Sin space left parenthesis x plus alpha right parenthesis times Sin space left parenthesis x minus alpha right parenthesis where 0 less than alpha less than pi for straight x in the interval  left square bracket 0 comma pi right square bracket is

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    The values of theta satisfying Sin space 5 theta equals Sin space 3 theta minus Sin space theta and 0 less than theta less than pi over 2 are

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    The value of k such that fraction numerator x minus 4 over denominator 1 end fraction equals fraction numerator y minus 2 over denominator 1 end fraction equals fraction numerator z minus k over denominator 2 end fraction lies in the plane 2x-4y+z+7=0 is

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    physics-General
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    Velocity-time open parentheses v minus t close parentheses graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is

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    maths-

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    x subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and x subscript 2 end subscript equals u t blank therefore x subscript 1 end subscript minus x subscript 2 end subscript equals fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent minus u t
    y equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent minus u t.This equation is of parabola
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    physics-General
    x subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and x subscript 2 end subscript equals u t blank therefore x subscript 1 end subscript minus x subscript 2 end subscript equals fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent minus u t
    y equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent minus u t.This equation is of parabola
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    As fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction greater than 0 i. e. comma graph shows possess minima at t equals fraction numerator u over denominator a end fraction
    General
    physics-

    A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in second which is as shownWhich one of the following plot represents velocity v in m s to the power of negative 1 end exponent blank v e r s u s time t in second?

    A particle starts from rest at t equals 0
    The equation of motion
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 blank m s to the power of negative 1 end exponent
    The velocity for next 2 s
    v to the power of ´ end exponent equals v plus a t
    blank equals 6 minus 3 cross times 2 equals 0
    v to the power of ´ end exponent equals 0

    Hence, v minus t graph will be as shown.

    A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in second which is as shownWhich one of the following plot represents velocity v in m s to the power of negative 1 end exponent blank v e r s u s time t in second?

    physics-General
    A particle starts from rest at t equals 0
    The equation of motion
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 blank m s to the power of negative 1 end exponent
    The velocity for next 2 s
    v to the power of ´ end exponent equals v plus a t
    blank equals 6 minus 3 cross times 2 equals 0
    v to the power of ´ end exponent equals 0

    Hence, v minus t graph will be as shown.