Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
Answer:
1848.15J
Explanation:
KE =1/2 mv^2
Mass = 60kg, velocity =40km/h =11.11m/s
Hence
KE =30 x(11.1)^2 /2 = 1848.15J
Answer:
<h2>
Work done by the gas is given as</h2><h2>
</h2>
Explanation:
As we know that the process is isothermal so here work done is given as
here we know that
now we have
so we have
Answer:
Explanation:
30 rev/min (2π rad/rev) / (60 s/min) = π rad/s
α = Δω/t = (0 - π)/3 = π/3 rad/s²
θ = ½αt² = ½(π/3)3² = 1.5π radians
θ = 1.5π rad/2π rad/rev = 0.75 rev
