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Svetllana [295]
3 years ago
14

A shaft 70 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250 mm long. T

he clearance, assumed uniform, is filled with oil with kinematic viscosity v = 0.005 m²/s and density p= 900 kg/m². Find the force exerted by the oil on the shaft.
Engineering
1 answer:
Allushta [10]3 years ago
6 0

Answer:

F=989.6 N

Explanation:

Given that

Diameter of shaft = 70 mm

Diameter of bearing sleeve =70.2 mm

So clearance h=0.1 mm

Speed V= 400 mm/s

Length of shaft = 250 mm

\nu =0.005\ \frac{m^2}{s}\ , \rho=900\ \frac{kg}{m^3}

We know that

\mu =\rho\times\nu

μ= 900 x 0.005 Pa-s

μ= 4.5 Pa-s

As we know that

From Newton's law of viscosity ,the shear stress given as follows

\tau =\mu \dfrac{dU}{dy}

We also know that

Force = shear stress x area

Now by putting the values

\tau =4.5\times \dfrac{400}{0.1}

\tau=18,000 Pa

So force

F= 18,000 x π x 0.07 x 0.25

F=988.6 N

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In a morphological matrix, the parameters that are essential for a design are in the left column.

<h3>What is a morphological matrix?</h3>

The morphological matrix is ​​a matrix where columns and rows represent the various parameters for solving a problem. The first column is used for the characteristics relevant to the problem; the horizontal lines are filled with possibilities for each of these parameters.

With this information, we can conclude that in a morphological matrix, the parameters that are essential for a project are in the left column.

Learn more about  morphological matrix in brainly.com/question/21120930

8 0
3 years ago
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo
agasfer [191]

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

8 0
3 years ago
For Figure Below, if the elevation of the benchmark A is 25.00 m above MSL:
nlexa [21]

Answer:

For Figure Below, if the elevation of the benchmark A is 25.00 m above MSL:

1. Using the Rise and Fall Method, find the reduced level for all points. (Construct the Table)

2. Using HPC Method, find the reduced level for all points. ( Construct the Table).

3. Show all required Arithmetic checks for your work. For Item 1 and 2.

4. What is the difference in height between points H and D?

5. What is the gradient of the line connecting A and J, knowing that horizontal distance = 200

m.

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4 years ago
Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician
mariarad [96]

Answer:

Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

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He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.

3 0
3 years ago
A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field
Assoli18 [71]

Answer:

=6.3*10^3 N/C

Explanation:

solution:

from this below equation (1)

E=σ/2εo(1-\frac{z}{\sqrt{z^2-R^2} } )...........(1)

we obtain:

=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )

=6.3*10^3 N/C

8 0
3 years ago
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