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Evgesh-ka [11]
2 years ago
12

Directions: Correct the errors.

Engineering
1 answer:
schepotkina [342]2 years ago
6 0

Answer:

1. I went to the library to study last night.

2.Helen borrowed my dictionary to look up the spelling of "occurred."

3.The teacher opened the window to let some fresh air into the room.

4..I came to this school to learn English.

5.I traveled to Osaka to visit my sister.

Explanation:

...

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Seperate real and imaginary parts tan(2x+i3y)
ahrayia [7]

Answer:

First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

3 0
3 years ago
Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1
WITCHER [35]

Answer:

Detailed solution is given below:

7 0
4 years ago
What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A
ikadub [295]

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

8 0
3 years ago
If a generator has 120 volts and 22 amps, how many ohms are needed so the generator will not explode.
daser333 [38]
120 volt divided by 22 ampere
= 5.4545454545455 ohm (Ω)
P = V × I
= 120 volt × 22 ampere
= 2640 watt (W)
7 0
3 years ago
An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

3 0
3 years ago
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