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Ostrovityanka [42]
3 years ago
6

Which sentence is an example of formal language?

Engineering
1 answer:
mote1985 [20]3 years ago
6 0
D sounds more formal than the rest.
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A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.
Rashid [163]

Answer:

1) Angle with x-axis = 42.03 degrees

2) Angle with y-axis =68.2 degrees

3) Angle with z-axis =   56.14 degrees

Explanation:

given any vector \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}

and any x axis the angle between them is given by

\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )

Applying values we get

\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}

Angle between the vector and y axis is given by

\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )

Applying values we get

\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}

Similarly angle between z axis and the vector is given by

\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )

Applying values we get

\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}

5 0
4 years ago
Which substance(s) have no fixed shape and no fixed volume?
Black_prince [1.1K]

Answer:

I think gas substances

5 0
4 years ago
Read 2 more answers
Manufacturers frequently make choices about their suppliers of raw materials based on their impact on society and the environmen
love history [14]
Manufacturers seek attention
4 0
3 years ago
The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv
Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$  in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$ \phi $ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              $ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

                              $= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

          = 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

                 $=\frac{1424500}{56.67 \pi}$

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

        $=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

      = 2.93 ksi

B). Angle of twist :

     $\phi = \frac{TL}{JG}$

         $=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

         = 0.08002 rad

         = 4.58°

6 0
3 years ago
Consider a piston-cylinder device with a piston surface area of 0.1 m^2 initially filled with 0.05 m^3 of saturated water vapor
miv72 [106K]

The friction force f = 10000 N

The heat transfer Q = 1.7936 KJ

<u>Explanation:</u>

Given data:

Surface area of Piston = 1 m^{2}

Volume of saturated water vapor = 100 K Pa

Steam volume = 0.05 m^{3}

Using the table of steam at 100 K pa

Steam density = 0.590 Kg/m^{3}

Specific heat C_{p} = 2.0267 KJ/Kg K

Mass of vapor = S × V

m = 0.590 × 0.05

m = 0.0295 Kg

Solution:

a) The friction force is calculated

Friction force = In the given situation, the force need to stuck the piston.

= pressure inside the cylinder × piston area

= 100 × 10^{3}  × 0.1

f = 10000 N

b)  To calculate heat transfer.

Heat transfer = Heat needs drop temperatures 30°C.

Q=m c_{p} \ DT

Q = 0.0295 × 2.0267 × 10^{3} × 30

Q = 1.7936 KJ

3 0
3 years ago
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