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3 years ago

10

Which is not required when working in a manufacturing facility?

1 answer:

Artyom0805 [142]3 years ago

5 0

Flip flops are not required

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A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang

AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

$\omega =\dfrac{d\theta}{dt}$

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

$V=\sqrt{\dfrac{GM}{R}}$

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

$V=\sqrt{\dfrac{GM}{R}}$

$V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}$

V=7059.44 m/s

iii)

We know that centripetal fore given as

$F=\dfrac{mV^2}{R}$

Here given that m= 200 kg

R= 8000 km

so now by putting the values

$F=\dfrac{mV^2}{R}$

$F=\dfrac{200\times 7059.44^2}{8000\times 10^3}$

F=1245.8 N

3 0

3 years ago

A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?

blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

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

9.19 Generate Bode magnitude and phase plots (straight-line approximations) for the following voltage transfer functions. (a) H(

Norma-Jean [14]

Answer:

attached below

Explanation:

8 0

3 years ago

An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th

makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)

v=total volume i.e. volume of container"80L" 80L=0.08 cubic meter

vf=vol of liquid phase vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0

3 years ago