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Physics: Post your doubts here!

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Q.16
When a ball is thrown up in da air, its P.E. increases and its K.E. decreases until it reaches maximum height, however da sum of K.E. n P.E. remains constant throught da flight. Answer: B

Q.17
V(x) = v V(y) = 2v
M(x) = 2m M(y) = 2

K.E of x = (1/2) * 2m * (v)² = mv²
K..E of y = (1/2) * m * (2v)² = (m/2) * 4v² = 2mv²

Therefore K.E. of y = 2 * K.E. of x
or K.E. of x = (1/2) * K.E. of y

Hence answer is : A

:)
 
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Q.16
When a ball is thrown up in da air, its P.E. increases and its K.E. decreases until it reaches maximum height, however da sum of K.E. n P.E. remains constant throught da flight. Answer: B

Q.17
V(x) = v V(y) = 2v
M(x) = 2m M(y) = 2

K.E of x = (1/2) * 2m * (v)² = mv²
K..E of y = (1/2) * m * (2v)² = (m/2) * 4v² = 2mv²

Therefore K.E. of y = 2 * K.E. of x
or K.E. of x = (1/2) * K.E. of y

Hence answer is : A

:)
Thanks :D
 
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Uncertainty question:

If R= 4.00±0.05, find R^2 including absolute uncertainty? i am not sure can anyone do it?
 
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Energy = Force * distance
E = Fs

When force 'F' is applied for a distance 's' , KE increases by 4 J or + 4 J
so
KE = F * s
Fs = 4 J . . . . . . (i)

When force '2F' is applied for a distance '2s', KE increases by 'x' , where x is unknown.
KE = 2F * 2s
4Fs = x
Substitute Fs from (i)
x = 4(4) J
[ x = 16 J ]

So KE increases by 16 J n becomes, 4 + 16 = 20 J

Answer: B :)
 
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