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zubka84 [21]
3 years ago
13

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)?

Physics
1 answer:
RSB [31]3 years ago
4 0

Answer:

b)  the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

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Determine the magnitude and direction of the resultant force of the following free body diagram.
Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force (\vec F), in newtons, by vectorial sum:

\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j} (1)

\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}

\|\vec F\| \approx 599.923\,N

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)

Where \theta is the direction of the resultant force, in sexagesimal degrees.

\theta \approx 36.405^{\circ}

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0
3 years ago
Select the correct answer.
lutik1710 [3]
I’m sorry i haven’t found the answer to this
8 0
3 years ago
Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the
masya89 [10]
The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2


8 0
4 years ago
How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?
Andrej [43]

Answer: B. 15.2s

Explanation: 600/39.5 = about 15.2

6 0
4 years ago
A driver is traveling eastward on a dirt road when she spots a pothole ahead. She slows her car from 14.0 m/s to 5.5 m/s in 6.0
Ivahew [28]

Answer:

- 1.42m/s²

Explanation:

Acceleration is defined as the change in velocity of a body with respect to time.

Acceleration = change in velocity/time

Change in velocity = final velocity - initial velocity

Acceleration = final velocity - initial velocity/time

Since she slows her car from 14.0 to 5.5m/s in 6seconds,

Initial velocity = 14m/s

Final velocity = 5.5m/s

Time = 6seconds

Substituting in the given formula, we will have

Acceleration = 5.5 - 14/6

Acceleration = - 8.5/6

Acceleration = - 1.42m/s²

The negative acceleration shows that the car decelerates.

8 0
3 years ago
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