We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
where
Now we find the moment of inertia by integrating from
to
The moment of inertia is
(from (-a/2) to
(a/2))
Answer:
Explanation:
Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree
From the formula of resultant vector
(1.703I)² = I² + I² + 2 I² cosФ
2.9 I² = 2I² + 2 I² cosФ
.9I² = 2 I² cosФ
cosФ = .9 / 2
= .45
Ф = 63.25 .
Answer:
3200°F
Explanation:
just add two zero's to the end
Answer:
Heat of vaporization will be 22.59 j
Explanation:
We have given mass m = 10 gram
And heat of vaporization L = 2.259 J/gram
We have to find the heat required to vaporize 10 gram mass
We know that heat of vaporization is given by , here m is mass and L is latent heat of vaporization.
So heat of vaporization Q will be = 10×2.259 = 22.59 J