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Anika [276]
3 years ago
11

What’s is 6 billion plus 6 quadrillions

Engineering
2 answers:
Leokris [45]3 years ago
8 0
12000000000000000000000000000000000
goblinko [34]3 years ago
3 0

Answer:

6.000006e+15 Or 6.000006000000000000000

Explanation:

I think pls brainliest me

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What is a p-n junction? Show by the diagram.
Natalija [7]

Answer:

The p-n junction is a region formed when a p -type semiconductor material is joined to an n-type semiconductor material

Explanation:

The p type semiconductor has holes as its majority charge carriers making it positively charged while the n –types has an overall negative charge. At the junction the holes move towards the electron until such a time when there is a balance in charges from both materials, which leads to the formation of the depletion zone as shown in the attachment below

                           

5 0
3 years ago
When would working with machinery be a common type of caught-in and caught-between<br> hazard?
tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between  hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

6 0
3 years ago
The critical resolved shear stress for a metal is 39 MPa. Determine the maximum possible yield strength (in MPa) for a single cr
damaskus [11]

Answer:

78 MPa

Explanation:

Given that the critical resolved shear stress for a metal is 39 MPa, the maximum possible yield strength for a single crystal of this metal is twice the critical resolved shear stress for the metal. The maximum yield yield strength for a single crystal of this metal that is pulled in tension (\sigma_y) is given as:

\sigma_y=2*critical\ resolved\ shear\ stress(\tau_{css})\\\\\sigma_y=2*\tau_{css}\\\\\sigma_y=2*39\\\\\sigma_y=78\ MPa

4 0
2 years ago
Technician A says that an A-pillar may be designed to transfer collision energy
svetoff [14.1K]
C both A and b cause they are technician both technicians so they both measure out the floor pan reinforcement be designed to transfer collision energy so I say both A and B
7 0
3 years ago
H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate
Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0 (1)

Where:

\dot m - Mass flow, in kilograms per second.

h_{in}, h_{out} - Specific enthalpies at inlet and outlet, in kilojoules per second.

v_{in}, v_{out} - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

p = 2000\,kPa

T = 300\,^{\circ}C

h_{in} = 3024.2\,\frac{kJ}{kg}

\nu_{in} = 0.12551\,\frac{m^{3}}{kg}

Where \nu_{in} is the specific volume of water at inlet, in cubic meters per kilogram.  

Outlet (Superheated steam)

p = 600\,kPa

T = 160\,^{\circ}C

h_{out} = 2758.9\,\frac{kJ}{kg}

If we know that \dot m = 50\,\frac{kJ}{kg}, h_{in} = 3024.2\,\frac{kJ}{kg}, h_{out} = 2758.9\,\frac{kJ}{kg} and v_{in} = 30\,\frac{m}{s}, then the flow speed at outlet is:

35765-25\cdot v_{out}^{2} = 0 (2)

v_{out} \approx 37.823\,\frac{m}{s}

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius (r_{in}), in meters, by this expression:

\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}} (3)

If we know that \dot m = 50\,\frac{kJ}{kg}, v_{in} = 30\,\frac{m}{s} and \nu_{in} = 0.12551\,\frac{m^{3}}{kg}, then the inlet radius is:

r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}

r_{in}\approx 0.258\,m

The inlet radius of the nozzle is approximately 0.258 meters.  

7 0
2 years ago
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