Select the correct answer.

For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5

bonufazy [111]

Answer:

The total force resisted by the flange bolts is 163.98 N

Explanation:

Solution

The first step is to find the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ = diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of the pipe at the inlet section

A₂ = π /4 D₂²

D₂ = diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265 * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0

3 years ago

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

V125BC [204]

Answer:

a) E = $\frac{K}{e*d}$ b) i.

Explanation:

Assuming no other forces acting on the electron, as the field is pushing the electron back till it comes to rest momentarily, applying the work-energy theorem, the loss in kinetic energy, must be equal to the work done by the field on the electron.

The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:

W = q*E = -e*E (1)

This work must be equal to the change in Kinetic Energy:

ΔK = Kf -K₀ = 0 - K = -K (2)

From (1) and (2), and solving for E, we have:

E = $\frac{K}{e*d}$

b) As the electric field (by convention) has the direction that it would have a positive test charge, the electrons (being negative charges),when starting from rest),have the direction opposite to the field.

In this case, as the electron has an initial velocity, and the field is opposing to the electron movement, we conclude that the electric field is in the direction of the electron´s motion, due to the electron slows down, and then comes momentarily to an stop, before changing direction and move opposite to the field as it is its natural behavior.

3 0

3 years ago

Your uncle has given you a newmonitor for your computer. When you attempt to connect it, you notice that none of the ports on th

juin [17]

Answer:

Go on amazon and look for HDMI -> (VGA/DVI/HDMI) what ever you need adapters.

6 0

3 years ago

The magic square is an arrangement of numbers in a square grid in such a way that the sum of the numbers in each row, and in eac

Yuki888 [10]

Answer:

See the explanation for the answer;

Explanation:

Matlab code is as given below

-------------------------------------------------------------------------------------Start of code

% Program to create a magic square and verify it

clc

M= magic(5)

% To find sum of elements of each row

r1= sum(M(1,:)) % Sum of row 1

r2= sum(M(2,:)) % Sum of row 2

r3= sum(M(3,:)) % Sum of row 3

r4= sum(M(4,:)) % Sum of row 4

r5= sum(M(5,:)) % Sum of row 5

% To find sum of each coloumn

c1= sum(M(:,1)) % Sum of coloumn 1

c2= sum(M(:,2)) % Sum of coloumn 2

c2= sum(M(:,3)) % Sum of coloumn 3

c2= sum(M(:,4)) % Sum of coloumn 4

c2= sum(M(:,5)) % Sum of coloumn 5

% To find sum of diagonal

d1= sum(diag(M)) % Sum of principal diagonal elements

d2= sum(diag(fliplr(M)))

-------------------------------------------------------------------------------End of code

Following results are obtained when executed.

M =

17 24 1 8 15

23 5 7 14 16

4 6 13 20 22

10 12 19 21 3

11 18 25 2 9

r1 =

65

r2 =

65

r3 =

65

r4 =

65

r5 =

65

c1 =

65

c2 =

65

c2 =

65

c2 =

65

c2 =

65

d1 =

65

d2 =

65

3 0

3 years ago

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y

kow [346]

Answer:

40π

Explanation:

First, find the limits (intersections).

5x² = 30x − 10x²

15x² − 30x = 0

x² − 2x = 0

x (x − 2) = 0

x = 0 or 2

Within this interval, 30x − 10x² is greater than 5x².

Dividing the volume into cylindrical shells, the volume of each shell is:

dV = 2π r h t

dV = 2π x (30x − 10x² − 5x²) dx

dV = 2π x (30x − 15x²) dx

dV = 30π (2x² − x³) dx

The total volume is the sum (integral):

V = ∫ dV

V = ∫₀² 30π (2x² − x³) dx

V = 30π ∫₀² (2x² − x³) dx

V = 30π (⅔ x³ − ¼ x⁴)|₀²

V = 30π (⅔ 8 − ¼ 16)

V = 30π (16/3 − 4)

V = 10π (16 − 12)

V = 40π

3 0

3 years ago