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]
Phosphorous is added to make an n-type silicon semiconductor with an electrical conductivity of 1.75 (Ωm)-1 . Calculate the necessary number of charge carriers required
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
CAD(computer-aided design) software and is used in__________and __________that show how to construct an object. Technical drawings show in detail how the pieces of something relate to each other.
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.
PLEASE ANSWER SOON
In a science lab, Cash mixes two clear liquids together in a beaker. Bubbles are produced, and a white solid forms and settles to the bottom. Which statement below describes what happened?
a
A physical change occurred, a gas and precipitate was produced
b
A physical change occurred, only a gas was produced
c
A chemical change occurred, only a gas was produced
d
A chemical change occurred, a gas and precipitate was produced
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.