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2 years ago

How do I do this? Blueprints, complete the missing view.

Engineering

1 answer:

Ymorist [56]2 years ago

7 0

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

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The viscosity of the water was 2.3×10^−5lb⋅⋅s/ft^2 and the water density was 1.94 slugs/ft^3. Estimate the drag on an 88-ft diam

Lina20 [59]

Answer:

hello your question is incomplete attached below is the complete question

answer : Drag force = 1.3 Ib

Explanation:

we have to represent the dimensions of the drag force in terms of FLT

i.e : D = f( <em>d,v,p,u </em>) represented in terms of FLT

D = F , V = LT^-1, d = L, p = FL^-4 T^2

u = FL^-2 T, Number of independent terms = 5

attached below is the detailed solution

6 0

3 years ago

Shear strain can be expressed in units of either degrees or radians. a)True b)- False

ExtremeBDS [4]

Answer:

true

Explanation:

shear strain is define as the ratio of change in deformation to the original length perpendicular to the axes of member due to shear stress.

ε = deformation/original length

strain is a unit less quantity but shear stain is generally expressed in radians but it can also be expressed in degree.

3 0

3 years ago

Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w

hoa [83]

Answer:

R = 4.75 lb (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.

8 0

2 years ago

(a) In a 3-phase, 4-wire system, the currents are in the A, B, and C lines under abnormal conditions of loading were as follows:

Verizon [17]

Answer:

Check the attached image below

Explanation:

Kindly check the attached image below to get the step by step explanation to the question above.

5 0

2 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

3 years ago