What cell type has no membrane-bound organelles, has DNA that is found in an area called the
nucleoid, and is very small?
A. Eukaryotic cell
B. Prokaryotic cell
C. Animal cell
D. Plant cell

Answer:nononononono

Explanation:

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: 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:

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]

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:

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

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:

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.