Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms
1 answer:
You might be interested in
Answer:
w = vR/3
Explanation:
The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.
From the principle of conservation of linear momentum , we have
m*v = 2*m* Vcm
Where v = velocity of bullet, Vcm = velocity of wood
Hence, we have
Vcm = v2
Also, from the conservation of angular momentum about the centre of mass.
M*V*(R/2) = Ic*w - equation (I)
where Ic = moment of inertia and w = angular velocity
Ic for a ring is given by
Ic of a bullet is given by
Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives
Ic(system) =
Substituting back into equation (I), we have
Hence, we obtain w =vR/3
w=v3R
Answer:
Explanation:
As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude
so we have
so now we have
now we have
so the phase other particle in opposite direction is given as
so we have phase difference given as