please hurry i’ll give you 15 points

Answer:nononononono

Explanation:

Answer:

15300 N

Explanation:

[tex]\rho_i[/tex] = Density of air at inlet

[tex]\dfrac{m}{t}[/tex] = Mass flow rate = 60 kg/s

[tex]v_i[/tex] = Inlet velocity = 225 m/s

[tex]\rho_o[/tex] = Density of gas at outlet = [tex]0.25\ \text{kg/m}^3[/tex]

[tex]A_i[/tex] = Inlet area

[tex]A_o[/tex] = Outlet area = [tex]0.5\ \text{m}^2[/tex]

Since mass flow rate is the same in the inlet and outlet we have

[tex]\rho_iv_iA_i=\rho_ov_oA_o\\\Rightarrow v_o=\dfrac{\dfrac{m}{t}}{\rho_oA_o}\\\Rightarrow v_o=\dfrac{60}{0.25\times 0.5}\\\Rightarrow v_o=480\ \text{m/s}[/tex]

Thrust is given by

[tex]F=\dfrac{m}{t}(v_o-v_i)\\\Rightarrow F=60\times (480-225)\\\Rightarrow F=15300\ \text{N}[/tex]

The thrust generated is 15300 N.

Answer:

Prokaryotic cell

Explanation:

Answer:

I think we would.

Explanation:

Only if u invent it.

answer

i don't think so

but it will take a lot of thinking if it is

possible

Answer:

I think it is D

Explanation:

I did a couple mins of research on this topic but there is no clear line that separates a physical and chemical reaction, so do with that as you will. i hope I somewhat helped.

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ([tex]\eta[/tex]), dimensionless, as:

[tex]\eta = \frac{\dot W_{out}}{\dot W_{in}}[/tex] (Eq. 1)

Where:

[tex]\dot W_{in}[/tex] - Maximum power possible from hydrogen flow, measured in kilowatts.

[tex]\dot W_{out}[/tex] - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

[tex]\dot W_{in} = \dot V\cdot \rho \cdot L_{c}[/tex] (Eq. 2)

Where:

[tex]\dot V[/tex] - Volume flow rate, measured in cubic meters per second.

[tex]\rho[/tex] - Density of hydrogen, measured in kilograms per cubic meter.

[tex]L_{c}[/tex] - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that [tex]\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}[/tex], [tex]\rho = 0.0899\,\frac{kg}{m^{3}}[/tex], [tex]L_{c} = 141790\,\frac{kJ}{kg}[/tex] and [tex]\dot W_{out} = 80\,kW[/tex], then the efficiency of this fuel cell is:

(Eq. 1)

[tex]\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)[/tex]

[tex]\dot W_{in} = 99.143\,kW[/tex]

(Eq. 2)

[tex]\eta = \frac{80\,kW}{99.143\,kW}[/tex]

[tex]\eta = 0.807[/tex]

The efficiency of this fuel cell is 80.69 percent.

Answer:

69

Explanation:

69420=420659782

fjkdod

Answer: A function returns a value and a procedure just executes commands.

Explanation:

Answer:

c - returns a value.

Explanation:

on edge pls mark brainiest

Answer:

The necessary number of electron charge carriers required is:

8.1019 × 10¹⁹ electrons/m³

Explanation:

The necessary number of charge carriers required can be determined from the resistivity. Given that, the phosphorus  make an n-type of silicon semiconductor;

Resistivity [tex]\rho = \dfrac{1}{\sigma}[/tex]

[tex]\rho = \dfrac{1}{q \mu _n n_n}[/tex]

where;

The number of electron on the charge carriers [tex]n_n[/tex] is unknown??

The charge of the electron q = [tex]1.6 \times 10^{-19} \ C[/tex]

The electron mobility  [tex]\mu_n[/tex] = 0.135 m²/V.s

The electrical conductivity [tex]\sigma[/tex] = 1.75 (Ωm)⁻¹

Making  [tex]n_n[/tex]  the subject from the above equation:

Then;

 [tex]n_n = \dfrac{\sigma }{q \mu_n}[/tex]

[tex]n_n = \dfrac{1.75 \ \Omega .m^{-1} }{1.6 \times 10^{-19} \times 0.135 \ m^2/V.s}[/tex]

[tex]n_n =8.1019 \times 10^{19}[/tex] electrons/m³

Thus; the necessary number of electron charge carriers required is:

8.1019 × 10¹⁹ electrons/m³

Answer:

Massive or linteled

Explanation:

Pyramid is a type of massive or linteled structure.

These structures do no have not much internal spaces and they are huge edifices.

A pyramid is a solid body with outer triangular faces that converges on top. To construct a pyramid, large amounts of materials are usually involved. Pyramids were more prominent in times past before this present civilization.

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

Answer:bbbb

Explanation:

Answer:

(166.0 mm Hg) (273.15 mL) = (760.0 mm Hg) (x)

Answer:

Plans; blueprints.

Explanation:

In Engineering, it is a common and standard practice to use drawings and models in the design and development of various tools or systems that are being used for proffering solutions to specific problems in different fields such as engineering, medicine, telecommunications and industries.

Hence, a design engineer make use of drawings such as pictorial drawings, sketches, or technical drawing to communicate ideas about a design to others, to record and retain informations (ideas) so that they're not forgotten and to analyze how different components of a design work together.

Technical drawing is mainly implemented with CAD (computer-aided design) software and is typically used in plans and blueprints that show how to construct an object.

Additionally, technical drawings show in detail how the pieces of something (object) relate to each other, as well as accurately illustrating the actual (true) shape and size of an object in the design and development process.

Answer:

1) 739 copies (739.7260273972603).

2) 5000 copies.

Explanation:

27000/365 = 73.97260273972603 (Their daily wins). How much times $.10 is equal to 73.97260273972603?

I used a two step equation to get it.

They need to sell 739.7260273972603 copies a day to get 27000$ a year.

an equation (y = mx+b) can be

y = 739.7260273972603x

To earn a 500$ profit we need to do the same two-step equation but instead of 73.97260273972603 use 500.

5000 copies needs to be sold to get a 500$ profit.

The last question I don't understand but I hope those 2 questions helped

Answer:

Create a google docs copy everything and paste hope this helps! :)

Explanation:

Answer:

Education and training I did the TEST and got it right.

Of the core elements of successful safety and health programs, the other one aside the given two that relates to individual roles is;

Education and Training Program.

According to OSHA(Occupational Safety and Health Administration), there are 7 core elements of an effective safety and health program.

These 7 core elements are namely;

1) Management Leadership.

2) Worker Participation.

3) Hazard Identification & Assessment.

4) Hazard Prevention & Control.

5) Education & Training Program.

6) Evaluation & Improvement.

7) Management of Contractors / Staffed Employees.

Now, we have been given 2 of the 7 which are Management leadership and

worker participation. However, we are told to find another one that relates

to individual roles. The only other one that is individual specific is Education

and Training program because it is the responsibility of every staff from top

to bottom to know all about safety.

Read more at; https://brainly.com/question/21555575

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

[tex]F = LwY_{avg}[/tex]

where;

w is the width of the annealed copper

[tex]Y_{avg}[/tex] is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

[tex]L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm[/tex]

Now, determine the average true stress, [tex]Y_{avg}[/tex], for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

[tex]F = LwY_{avg}[/tex]

[tex]F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN[/tex]

The power required in this operation is given by;

[tex]P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW[/tex]