Diverging lens=concave lens
The image would be after the first focal point but before the lens and it will be smaller than the object. Look at my picture for reference. Please let me know if this helped you!
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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The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing for n and 202Hz for
The frequency of the tightened is 205Hz
Answer:
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
Explanation:
a) The mass flow rate through the nozzle can be calculated with the following equation:
Where:
: is the initial velocity = 20 m/s
: is the inlet area of the nozzle = 60 cm²
: is the density of entrance = 2.21 kg/m³
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
Therefore, the exit area of the nozzle is 23.6 cm².
I hope it helps you!