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3 years ago

Nanotechnology is a scientific area that deals with making or changing things that are incredibly _______________. *

2 answers:

8 0

Answer:

B - Tiny

Explanation:

When you’re dealing with nanotechnology, you’re dealing with technology conducted at the nanoscale, which is about 1-100 nanometers. Nanometers have a length of one billionth of a meter. So options A and C would be incorrect. Option D wouldn’t necessarily be a good choice either.

Dafna1 [17]3 years ago

3 0

B tiny nano means small!!

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The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the air

navik [9.2K]

Answer:

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1 Subsonic

Mach > 1 Supersonic

Ma= Speed of the object/Speed of the sound

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

8 0

3 years ago

A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

D=200 mm

P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

hoop stress $\sigma _{h}=\frac{Pd}{2t}$

Longitudinal stress $\sigma _{l}=\frac{Pd}{4t}$

So maximum shear tress in plane $\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}$

$\tau _{max}=\dfrac{Pd}{8t}$

Now by putting the value

$27.5=\dfrac{4\times 200}{8t}$

So t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0

3 years ago

A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst

svetoff [14.1K]

Answer:

a) fuel cell weight and volume: 50 kg, 33.3 L

b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

Fuel Cell

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

(50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

(50 kW)/(1 kW/kg) = 50 kg

Fuel Tank

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

(50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

volume = 2571.43 MJ/(4 MJ/L) = 642.9 L

mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

_____

Comment on fuel tank volume

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

4 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.

change in mechanical energy is givrn as

$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

= 0.491 kJ/kg

$\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW$

$\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76$

B) $\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80$

c) $\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)$

$\dot W_{shaft} = 1964 kW$

7 0

3 years ago

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

2 years ago