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saveliy_v [14]
3 years ago
5

How did humans create a space suit without ever going. How did we know spaces conditions?

Engineering
2 answers:
brilliants [131]3 years ago
6 0

Answer:

ser gay

Explanation:

no es malo y se fjfjdj gracias

saul85 [17]3 years ago
5 0
Wiley Post experimented with a number of pressure suits for record-breaking flights. Russell Colley created the space suits worn by the Project Mercury astronauts, including fitting Alan Shepard for his ride as America's first man in space
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In a photonic material, signal transmission occurs by which of the following? a)- Electrons b)- Photons
fredd [130]

B. Photons.

In a photonic material, signal transmission occurs by photons which are light particles.

8 0
3 years ago
Problem 4.041 SI Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -26oC with a vo
Rom4ik [11]

Answer:

0.0297M^3/s

W=68.48kW

Explanation:

Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

state 1

X=quality=1

T=-26C

density 1=α1=5.27kg/m^3  

entalpy1=h1=234.7KJ/kg

state 2

T2=70

P2=8bar=800kPa

density 2=α2=31.91kg/m^3  

entalpy2=h2=306.9KJ/kg

Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.

m1=m2

(Q1)(α1)=(Q2)(α2)

\frac{(Q1)(\alpha 1) }{\alpha 2} =Q2\\Q2=\frac{(0.18)(5.27) }{31.91} =0.0297M^3/s

the volumetric flow rate at the exit is 0.0297M^3/s

To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation

W=m(h2-h1)

m=Qα

W=(0.18)(5.27)(306.9-234.7)

W=68.48kW

the compressor power is 68.48kW

4 0
3 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
2. BCD uses 6 bits to represent a symbol. a) True b) False​
Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0
3 years ago
DRIVERS ED
forsale [732]

Answer:

b

Explanation:

only if there signal is turned on

8 0
3 years ago
Read 2 more answers
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