Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by
![P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7BT%5Cmu%7D%5Comega%5E2y%5E2%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B1871%5Ctimes%203.9%5Ctimes%2010%5E%7B-3%7D%7D%5Ctimes%201203%5E2%5Ctimes%20%283.1%5Ctimes%2010%5E%7B-3%7D%29%5E2%5C%5C%5CRightarrow%20P%3D18.7842493212%5C%20W)
The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
Answer:
θ= 5 radian
Explanation:
Given data:
Radius r = 0.70 m
Initial angular speed ω_i = 2rev/s
Time t = 5 s
Final angular speed ω_f =0
so we have angular displacement
![\theta= \frac{\omega_f-\omega_i}{2}\times t](https://tex.z-dn.net/?f=%5Ctheta%3D%20%5Cfrac%7B%5Comega_f-%5Comega_i%7D%7B2%7D%5Ctimes%20t)
putting values
= 5 rad
Answer:
Moment of inertia of the solid sphere:
I
s
=
2
5
M
R
2
.
.
.
.
.
.
.
.
.
.
.
(
1
)
Is=25MR2...........(1)
Here, the mass of the sphere is
M
M
Answer:
S = 1/2 Vo t + 1/2 a t^2 = d time for particle to travel distance d
F = E q force acting on particle
a = F / m = E q / m
d = Vo t + E q / (2 m) t^2
One would need to solve the quadratic equation shown to find the time t
t^2 + (2 m) / E q * V0 t - (2 m) / E q * d = 0
or t^2 + A V0 t - A d = 0 where A = (2 m) / E q