Answer:
A. 0.66 Amps
Explanation:
Using ohms law, we can say that Voltage is equivalent to Current times Resistance. We are given the voltage and the resistance of the circuit, so we simply need to find the current.
V = IR
Solve for I, where V = 2volts and R = 3ohms.
V = IR
V * 1/R = I * R * 1/R
I = V/R
I = 2/3 Amps
Hence, we should choose option A, 0.66 Amps for the current in this simple circuit.
Cheers.
Which of the following answer options can damage flexible and extension cords?
Staples
Door or window
Fastenings
Aging
Abrasion from adjacent
(Select all that apply)
Answer:
staples
Door or window
Explanation:
staples.- can hit the internal core and risk of electrocution.
Door or window - can damage the outer core
staples.- can hit the internal core and risk of electrocution or Door or window - can damage the outer core.
What is Extension cords?A length of flexible electrical power cable (flex) with a plug on one end and one or more sockets on the other end (often of the same kind as the plug) is known as an extension cord (US), power extender, drop cord, or extension lead (UK).
The phrase is also used to describe extensions for various types of cabling, but it mainly refers to mains (home AC) extensions.
The phrase "adapter cord" may be used if the plug and power outlet are of different types. Although they can be produced up to 300 feet long, most extension cords are between two and thirty feet long.
Therefore, staples.- can hit the internal core and risk of electrocution or Door or window - can damage the outer core.
To learn more about Extension cords, refer to the link:
https://brainly.com/question/14925233
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Below are listed the atomic weight, density and atomic radius for three hypothetical metal alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.
Alloy
A
B
C
Atomic
Weight
(g/mol)
43.1
184.4
91.6
Density
(g/cm
3)
6.40
12.30
9.60
Atomic
Radius
(nm)
0.122
0.146
0.137
The names of each of the crystal structures for each of the given metal alloys are; Alloy A is BCC structure; Alloy B is SC structure; Alloy C is BCC structure
What is a Crystal Structure?To solve for each of the three given alloys, we will need to do so by trial and error method to calculate the density and compare it to the value given in the problem.
The density formula is;
ρ = nA/(V_c × N).
Where;
n is 1, 2 and 4 for SC, BCC & FCC crystal structures respectively.
A is atomic weight
V_c is volume expressed as a³.
a is given by 2R,4R/√3, 2R√2 for SC, BCC & FCC crystal structures respectively with R being the atomic radius.
N is avogadros number of atoms and has a constant value of 6.022 × 10^(23) atoms/mol
A) For metal alloy A, we are given;A = 43.1 g/mol
If we assume it's a BCC structure, then;
n = 2
V_c = a³ = (4R/√3)³ = (4 × 1.22 × 10^(-8) × 1/√3)³ = 22.4756 × 10^(-24) cm³
N = 6.022 × 10^(23) atoms/mol
Plugging in the relevant values into the density formula, we have;
ρ = (2 × 43.1)/(22.4756 × 10^(-24) × 6.022 × 10^(23)) ≈ 6.4 g/cm³
This value is same as the one in the question, thus metal alloy A is a BCC crystal structure
B) For metal alloy B, we are given;A = 184.4 g/mol
If we assume it's a SC structure, then;
n = 1
Thus;
V_c = a³ = (2 × 1.46 × 10^(-8))³ = 24.897 × 10^(-24) cm³
Plugging in the relevant values into the density formula, we have;
ρ = (1 × 184.4)/(24.897 × 10^(-24) × 6.022 × 10^(23)) ≈ 12.3 g/cm³
This value is same as the one in the question, thus metal alloy B is an SC crystal structure
C) For metal alloy C, we are given;A = 91.6 g/mol
If we assume it's a BCC structure, then;
n = 2
V_c = a³ = (4R/√3)³ = (4 × 1.37 × 10^(-8) × 1/√3)³ = 31.67 × 10^(-24) cm³
Plugging in the relevant values into the density formula, we have;
ρ = (2 × 91.6)/(31.67 × 10^(-24) × 6.022 × 10^(23)) ≈ 9.6 g/cm³
This value is same as the one in the question, thus metal alloy C is an SC crystal structure.
Read more about crystal structures at;https://brainly.com/question/14831455
How is the caliber of a rifle or handgun determined
A data bus can be visualized as a multilane highway
A. and each component is located at an intersection where it will turn or go straight
B. with each component having an individual address
C. and each component located in a curve where the data will slow down
D. with each component acting as a traffic light, stop-go
Answer:
B. with each component having an individual address
Explanation:
A data bus is a system within a computing system, that consists of a set of wires or connectors, that provides transportation for data. A data bus can transfer data and information through a computing system, or between two or more computing systems. Each component in the computer has its own unique address by which data or information is sent to it, to or fro the central processing unit, and the data or information travelling through the data bus, which serves as the highway, reaches each component using this address. This is analogous to the postal service system of moving letters and packages, through a series of networks that locates the address of the receiver.
The subsequent result of the
system to the input is known
as
Select one:
A.Response
B.Command
C.Process Control
D.Process Controller
as
A rectangular channel 3 m wide carries 4 m3/s of water at a depth of 1.5 m. If an obstruction 15 cm high is placed across the channel, calculate the elevation of the water surface over the obstruction. What is the maximum height of the obstruction that will not cause a rise in the water surface upstream
Answer:
the elevation of the water surface over the obstruction is highest at 1.341 m
the maximum height of the obstruction h = 0.159 m
Explanation:
From the given information:
The diagrammatic expression for the water profile showing a rectangular channel with 3 cm width carrying a 4 m3/s of water at a depth of 1.5 m with an obstruction 15 cm high is placed across the channel can be seen in the diagram attached below.
To calculate the elevation of the water surface over the obstruction, we need to determine the following:
a. the velocity of the channel
b. the froude number at the upstream of the obstruction
c. the specific energy level
To start with the velocity V of the channel.
[tex]V_1 = \dfrac{Q}{A} \\ \\ V_1 = \dfrac{4 \ m^3/s}{(3 \times 1.5 ) m^2 } \\ \\ V_1 = \dfrac{4 \ m^3/s}{4.5 \ m^2}[/tex]
[tex]V_1 = 0.88 \ m/sec[/tex]
The froude number at the upstream of the obstruction
[tex]F _{\zeta} = \dfrac{V_1}{\sqrt{gy__1}}[/tex]
[tex]F _{\zeta} = \dfrac{0.88}{\sqrt{9.81 \times 1.5}}[/tex]
[tex]F _{\zeta} = \dfrac{0.88}{\sqrt{14.715}}[/tex]
[tex]F _{\zeta} = \dfrac{0.88}{3.836}[/tex]
[tex]F _{\zeta} =0.229[/tex]
[tex]F _{\zeta} \simeq0.3[/tex] which is less than the subcritical flow.
Similarly, the specific energy level for this process can be expressed as:
[tex]E_1 = \dfrac{V_1^2}{2g}+y_1[/tex]
[tex]E_1 = \dfrac{0.88^2}{2 \times 9.81 }+1.5[/tex]
[tex]E_1 = \dfrac{0.7744}{19.62 }+1.5[/tex]
[tex]E_1 =0.039469+1.5[/tex]
[tex]E_1 =1.539469[/tex]
[tex]E_1 \simeq1.54 \ m[/tex]
[tex]E_2 + \Delta z = E_1[/tex]
[tex]E_2= E_1 - \Delta z[/tex]
[tex]E_2= (1.54 - 0.15) \ m[/tex]
[tex]E_2=1.39 \ m[/tex]
Suppose ;
[tex]V_1 y_1 = V_2y_2\\ \\ Then; \ making \ V_2 \ the \ subject \ of \ the \formula\ we \ have: \\ \\ \\V_2 = \dfrac{V_1 y_1}{y_2} ---- (1)[/tex]
From the energy equation:
[tex]E_2 + \Delta z = E_1[/tex]
we can now substitute the above derived parameter and have :
[tex]E_2 + \Delta z=y_1 + \dfrac{V_1^2}{2g}[/tex]
[tex]E_2 =y_1 + \dfrac{V_1^2}{2g} - \Delta z[/tex]
[tex]E_2 =y_2 + \dfrac{V_2^2}{2g}[/tex]
replace the value of [tex]V_2[/tex] =[tex]\dfrac{V_1 y_1}{y_2}[/tex] in equation (1), we have:
[tex]E_2 =y_2 + \dfrac{( \dfrac{V_1y_1}{y_2})^2}{2g}[/tex]
[tex]E_2 =y_2 + \dfrac{ {V_1^2y_1^2}}{2gy_2^2}[/tex]
[tex]E_2 \times y_2^2 =y^3_2 + \dfrac{ {V_1^2y_1^2}}{2g}[/tex]
[tex]y^3_2 - E_2y^2_2 + \dfrac{V_1^2y_1^2}{2g}=0[/tex]
Replacing our values now; we have:
[tex]y^3_2 - 1.39 \times y^2_2 + \dfrac{0.88^2 \times 1.5^2}{2 \times 9.81}=0[/tex]
[tex]y^3_2 - 1.39 \times y^2_2 + \dfrac{1.7424}{19.62}=0[/tex]
[tex]y^3_2 - 1.39 \times y^2_2 + 0.0888=0[/tex]
[tex]y^3_2 - 1.39 y^2_2 + 0.0888=0[/tex]
[tex]y_2 = 1.341 \ m \\ \\ y_2 = - 0.232 \ m \\ \\ y_2 = 0.281 \ m[/tex]
Therefore,the elevation of the water surface over the obstruction is highest at 1.341 m
What is the maximum height of the obstruction that will not cause a rise in the water surface upstream
In order to determine the maximum height , we need to first estimate the rise in water level surface of [tex]\Delta z[/tex]
[tex]\Delta z[/tex] = 1.5 - (1.341+ 0.15) m
[tex]\Delta z[/tex] = (1.5 - 1.491) m
[tex]\Delta z[/tex] = 0.009 m
Finally, the maximum height of the obstruction h = (0.009 + 0.15 )m
the maximum height of the obstruction h = 0.159 m
Which of the following has led to a safer and more prosperous global community within the last century? the Bronze Age composite materials new materials ceramics
Answer:
The bronze age
Answer: New materials
Explanation: I read it in the article.