A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation
rate) ωi, and a stationary outer cylinder of inside radius Ro. In the tiny gap between the two cylinders is the fluid of viscosity μ. The length of the cylinders (into the page) is L. L is large such that end effects are negligible (we can treat this as a two-dimensional problem). Torque (T) is required to rotate the inner cylinder at a constant speed. (a) Showing all of your work and algebra, generate an approximate expression for T as a function of the other variables.
(b) Explain why your solution is only an approximation. In particular, do you expect the velocity prole in the gap to remain linear as the gap becomes larger and larger (i.e., if the outer radiusR0 were to increase, all else staying the same)?
(b) Explain why your solution is only an approximation. In particular, do you expect the velocity prole in the gap to remain linear as the gap becomes larger and larger (i.e., if the outer radiusR0 were to increase, all else staying the same)?
1 answer:
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Answer:
so here it will move in circle with radius 4.06 cm
Explanation:
As we know that proton is moving towards west while the magnetic field is vertically upwards
So here the force on the proton must be perpendicular to the velocity
So here we have
so here we have
since force is perpendicular to the velocity so here it must be centripetal force
here we have
so we have
so here it will move in circle with radius 4.06 cm
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒
(b)
For m₃,
⇒
Or,
⇒
Answer:
a) , b)
Explanation:
a) The magnetic force experimented by a particle has the following vectorial form:
The charge of the electron is equal to . Then, cross product can be solved by using determinants:
The magnetic force is:
b) The charge of the proton is equal to . Then, cross product has the following determinant:
The magnetic force is:
Answer:
a) W =400 kJ
b) W = 0 kJ
c) W =-160.944 KJ
Explanation:
<u>Given </u>
<u><em>Process 1 ---> 2 </em></u>
The relation of the process P = constant
Pressure of point (1) P1 = 10 bar = P2
Volume of point (1) V1 = 1 m^3
Volume of point (2) V2 =4 m^3
The relation of the process V = constant
<u>Process 2 ---> 3 </u>
The relation of the process V = constant
V3 = V2
Pressure of point (3) P3 = 10 bar
Volume of point (3) V3 = 4 m^3
<u>Process 3 ---> 1 </u>
The relation of the process PV = constant
<u>Required </u>
Sketch the processes on the PV coordinates
The work for each process in kJ
<u>Solution </u>
The work is defined by
W=
<em>a=V2</em>
<em>b=V1</em>
<em>x=P</em>
<em>dx=dV</em>
<u>Process 1 ---> 2 </u>
P3 = P4 = 5 bar
W=
<em>a=V3</em>
<em>b=V2</em>
<em>x=4</em>
<em>dx=dV</em>
putting the value of a, b, x, dx in above integral
W=400 kJ
<u>Process 2 ---> 3 </u>
V = constant Then there is no change in the volume,hence W = 0 kJ
<u>Process 3 ---> 1 </u>
By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1
W=
a=V1
b=V3
x=1V^-1
dx=dV
putting the value of a, b, x, dx in above integral
W=| ln V | limit a and b
= -160.944 KJ
