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Comparison of Rational Numbers

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  • Volume

    Q

    The buoyancy is 450-310 g = 140g
    So the body displaces the volume of that 140g of water.
    Which is 140 cm^3
    because water weighs 1 g per 1 cm^3

    A
  • Rectilinear Figures

    Q

    Solution:-
    For drawing a parrallelogram we can find out the length of the base of the perpendicular from the vertex B of the parallelogram.
    In triangle AOB
    AO2 + BO2 = AB2
    BO2 = (4.5)2 - (3.5)2

    BO2 = 8
    BO = 2.83 cm

    A
  • Digestive System

    Q

    Altogether, about 5% of starch digestion happens in the mouth. However, it will vary a bit depending on how long you chew, how much moisture there is in the mouth, and how much moisture the starch can absorb.

    A
  • Retradation

    Q

    Retardation is negative acceleration and defined as rate of decreasing of velocity of an object. It’s SI unit is m/s².

    A
  • Mid - Point Theorems

    Q

    Let PQRS� be a rectangle and ABCD the quadrilateral whose midpoints form the rectangle�
    P is midpoint of AB �
    Q is midpoint of BC �
    R� is midpoint�of CD
    S is midpoint of DA
    PS divides triangle ABD in the proportion 1:2�
    All 4 sides of the rectangle divide the corresponding sides of the triangles formed in a similar manner.�
    Similarly, sides of the rectangle are parallel to the diagonals of the quadrilaterals respectivly.
    Lines perpendicular to the same line are perpendicular.
    Therefore, AC is perpendicular to BC.

    A
  • The Language of Chemistry

    Q


    A
  • Indices

    Q

    (x+y+z) / (x^(-1) y^(-1) + y^(-1) z^(-1) + z^(-1) x^(-1) )�
    = (x+y+z) / (1/xy + 1/yz + 1/xz )�
    = (x + y + z)/{(x + y + z)/(xyz)}
    = xyz

    A
  • Newton's Law Of Motion

    Q

    A cyclist riding along a level road does not come to rest immediately after he stops pedaling.�Becausethe bicycle continues to move due to inertia of motion even after the cyclist stops pedaling i.e., applying the force. But it comes to rest afterwards as a result of the retarding force of friction between the tyres of bicycle and the ground.

    A
  • Newton's Laws Of Motion

    Q

    DOWNWARD MOTION:
    FInd the final velocity just before touching the ground.
    v1 = - sqrt(2gh1) where h1 = 40m
    v1 = -28 m/s (take care of negative sign)
    UPWARD MOTION:
    FInd the initial velocity:
    u1 = sqrt(2gh)
    u1 = 14m/s
    Therefoe, J = m(u1-v1) = 0.75(14 - (-28)) = 31.5
    Note: Take care of sign

    A

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