A pipe branches symmetrically into two legs of length L, and the whole system rotates with angular speed ω around its axis of sy
mmetry. Each branch is inclined at angle α to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate Q. The pipe diameter, D, is much smaller than L. Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration ω_ ?
1 answer:
You might be interested in
Answer:
1. 18.25 m/s
2. 0 m/s
Explanation:
1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account
The speed at this point would then be
2. Similarly, if T = mg, then the centripetal acceleration must be
As the ball has no centripetal acceleration, its speed must also be 0 as well.
Explanation:
It is given that,
Wavelength of x-rays = 2 nm
Plane spacing, d = 0.281 nm
It is assumed to find the scattering angle for second order maxima.
For 2nd order, Bragg's law is given by :
For second order, n = 2
Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.