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Kamila [148]
3 years ago
7

PLEASE HELP ASAP.

Physics
1 answer:
kifflom [539]3 years ago
4 0

Volume of the original block (V) of length (l), width (w) and height (h) is given by the following equation

V = l*b*h

hence volume of original block, V =  (6)(4)(10) = 240 cm^{3}

Now say we cut this block in two equal parts, than the length of new block will    be 6/2 = 3 cm ; where as its width and height will be same as of original block i.e. 4 cm and 10 cm respectively.

So, Volume of half-block is = (3)(4)(10) = 120 cm^{3} ; or we can say that the volume of half-block is half that of orignal block.

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Two billiard balls with the same mass undergo a perfectly elastic head-on collision. if one ball's initial speed was 2 m/s, and
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<span>The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
 The ball with an initial velocity of 3.6 m/s rebounds at 2 m/s

   There are two principles involved here
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  I'll use the following variables
 a0, a1 = velocity of ball a (before and after collision)
  b0, b1 = velocity of ball b (before and after collision)
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   For conservation of momentum, we can create this equation:
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   The equation a1^2 + b1^2 = 16.96 describes a circle centered at the origin with a radius of sqrt(16.96). The equation a1 + b1 = -1.6 describes a line with slope -1 that intersects the circle at two points. Those points being (a1,b1) = (-3.6, 2) or (2, -3.6). This is not a surprise given the conservation of energy and momentum. We can't use the solution of (2, -3.6) since those were the initial values and that would imply the 2 billiard balls passing through each other which is physically impossible. So the correct solution is (-3.6, 2) which indicates that the ball going 2 m/s initially rebounds in the opposite direction at 3.6 m/s and the ball originally going 3.6 m/s rebounds in the opposite direction at 2 m/s.</span>
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