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2 years ago

Which items are NOT found on a

Engineering

1 answer:

Alja [10]2 years ago

3 0

Answer:

None of the above cause thats what i put

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Which career best fits the group of the words below? High voltage, lines

Llana [10]

Answer:

I forget the word for it, but probably the guys who set up the power lines in the city.

Explanation:

5 0

2 years ago

Read 2 more answers

Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Repor

REY [17]

Answer:

$T_2=315.69k$

Explanation:

Initial Temperature $T_1=500K$

Initial Pressure $P_1=1000kPa$

Final Pressure $P_2=200kPa$

Generally the gas equation is mathematically given by

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}$

Where

n for $CO=1.4$

Therefore

$\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}$

$T_2=315.69k$

7 0

2 years ago

Hey answr this sajida Yusof

Allisa [31]

What do u need? Rusbaisuwvwbs

5 0

2 years ago

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Problem 2: Sieve Analysis and Soil Gradation

Marina CMI [18]

<u>Explanation:</u>

% Gravel = 100 - 72% = 28%

% Sand = 100 - 28 - 15 = 57%

% Fires = 100 - 0 - 28 - 57 = 15%

$\begin{aligned}&D_{n} \cdot 0.03 \quad ; \quad D_{x}=1.1 \quad ; \quad D_{0}=3.3\\&C_{u}=\frac{D_{60}}{D_{10}}=\frac{3.3}{0.03}=110\\&C_{c}=\frac{D_{30}^{2}}{D_{0} \cdot D_{0}}=\frac{1.1^{2}}{3.3 \times 0.03}=12.22\end{aligned}$

The soil is poorly graded soil.

6 0

2 years ago

**Please Help, ASAP**

Novay_Z [31]

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

$x=\dfrac{y}{1+y}$

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

$x=\dfrac{-y}{1-y}$

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

$x=\dfrac{y}{1-y}$

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

$2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}$

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

$\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}$

(6) ay^2 = x^3, (y)

$y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}$

Hence, this is the required solution.

3 0

2 years ago