Click and drag on elements in order
Rank the following species in order of increasing strength to act as an oxidizing agent from weakest at the top of the list to strongest at the bottom of the list.
K+(aq) + e- → K(s); Eo = -2.925 V
Hg2+(aq) + 2e- → Hg(l); Eo = +0.854 V
I2(s) + 2e- → 2I-(aq); Eo = +0.536 V

The unknown capacitor in arm CD must have a value of 330 microfarads, and the unknown resistor must have a value of 100 kilo-ohms.

To solve the problem, use the following formula:
Cseries = C1 x C2 / (C1 + C2)

Where C1 is the value of the capacitor in arm AB (47 microfarads) and C2 is the value of the capacitor in arm BC (330 microfarads).

Therefore, Cseries = 330 microfarads.

Also, the total resistance of arms CD is the sum of the resistance of the resistor (R) and the reciprocal of the capacitive reactance of the capacitor (1/Xc).

Using the following formula:
Rtotal = R + 1/Xc

Where Xc = 1/2πfC,

f is the frequency and C is the capacitance.

For this problem,

Xc = 1/2π(50)(330 x 10-6)

=> 100 kilo-ohm.

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The resultant vector has a magnitude of 77.3 N and is traveling at an angle of 54.3 degrees with the x-axis.

To solve the problem, use the Pythagorean theorem and trigonometric functions. Use the given values of the forces to find the resultant vector and the angle at which the vector is traveling. A resultant vector is the sum of two or more vectors.

In physics, force vectors are used to represent forces acting on objects. Forces are represented by vectors because they have both magnitude and direction. Therefore, when two or more forces are applied to an object, the resultant force can be found by adding up the individual forces.

The given forces are 50 N at 45° and 30 N at 70°. Let F₁ be the first force with a magnitude of 50 N at an angle of 45 ° with the x-axis. Let F₂ be the second force with a magnitude of 30 N at an angle of 70° with the x-axis.

To find the resultant vector, we need to find the x and y components of each force. To do this, use trigonometric functions.

For example, the x-component of F₁ can be found using cosine: F₁x = 50 cos(45°) = 35.4 N.

The y-component of F₁ can be found using sine:

F₁y = 50 sin(45°) = 35.4 N.

Do the same for F₂:

F₂x = 30 cos(70°) = 9.48 N, and F₂y = 30 sin(70°) = 28.7 N.

The x-components add up to give the total x-component:

Fx = F₁x + F₂x = 35.4 N + 9.48 N = 44.9 N.

The y-components add up to give the total y-component:

Fy = F₁y + F₂y = 35.4 N + 28.7 N = 64.1 N.

Use the Pythagorean theorem to find the magnitude of the resultant vector:

F = √(Fx² + Fy²) = √((44.9 N)² + (64.1 N)²) = 77.3 N.

Use trigonometric functions to find the angle of the resultant vector:

tanθ = Fy/Fx = 64.1 N/44.9 N.

Therefore, θ = tan⁻¹(64.1/44.9) = 54.3°.

The resultant vector has a magnitude of 77.3 N and is traveling at an angle of 54.3 degrees with the x-axis.

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The rate laws corresponding to the surface reactions as the rate-limiting step in the liquid phase hydration of isobutene has to be determined.

The rate law of the chemical reaction states that the rate of reaction is the function of the concentration of the reactants and the products present in that specific reaction. The rate is actually predicted by the slowest step of the reaction.

If there is a chemical reaction which has reactants A and B that reacts to form products then their rate law is given as follows.

   r=k[A]a[B]b

Here, [A] is the concentration of the reactant A, [B] is the concentration of the reactant B and k is the rate constant.

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Answer:

Milk Production, Health, and Age.

Explanation:

You want a younger cow when buying cattle so you can have gather more milk from it over it's lifetime. You also want to make sure that it can actually produce milk. Then you want a cow in good health.

In regards to discussing automotive batteries. Both technician "A" and "B" are correct. Hence option B is correct.

Automotive batteries must be disposed of or recycled at dedicated facilities because they contain lead and acid, which are hazardous materials. These facilities are equipped to properly handle and dispose of these materials in an environmentally safe way.

Therefore, When jump starting a vehicle, the last connection should be the positive cable. This is because the positive cable carries the electrical charge that starts the vehicle, and connecting it last prevents any sparks from occurring near the battery, which could cause an explosion or fire.

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Answer:

  36π inches ≈ 113.0973 inches

Explanation:

The circumference of the wheel is pi times its diameter.

  C = πd

  C = π(36 in) = 36π in ≈ 113.0973 in

The distance around the wheel is 36π inches, about 113.0973 inches.

Answer:

Uh- dude if you think I'm gonna download that, think twice...

Answer:

hard bored

Explanation:

it seems like it could be either but I think hard bored

The duty cycle should the converter have is the stack is delivering power to a 48 volt DC bus is 0.541.

The correct answer is option c. 0.541.

In order to control the output voltage of a DC/DC converter, a boost converter uses a pulse-width-modulated square wave with a variable duty cycle. To obtain a higher voltage than the input voltage, this control element is used.

The formula for calculating duty cycle is:Duty Cycle = Vout/Vin

Where Vout is the output voltage and Vin is the input voltage.

The power delivered by the fuel cell stack is:

P = V x I

where V is voltage and I is current given. Therefore, P = 25V x 39A = 975 watts.

The input power to the boost converter is equal to the output power of the fuel cell:

P in = P out975

W = V out x I out

V out = 48 volts

I out = 975/48 = 20.3 amps

Now, the duty cycle can be calculated:Duty Cycle = Vout/Vin= 48/25= 1.92

Therefore, Duty Cycle = 1/(1+D) => 1.92 = 1/(1+D)D = 0.521

So, the duty cycle should be 0.541 (option c).

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Formwork is a temporary structure that is used in construction to provide support and shape to freshly poured concrete until it is strong enough to stand on its own.

It is a crucial component of any construction project as it ensures that the concrete is poured into the desired shape and size, and prevents any potential collapses or deformations. Additionally, formwork helps to create a smooth and uniform finish to the concrete surface, which is important for both aesthetic and functional purposes. Overall, formwork is an essential tool in modern construction, providing a safe and reliable method of shaping and supporting concrete structures.

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We can decompose the supplied pseudoinstruction into smaller parts and implement it with a minimal set of MIPS instructions.

Here is one conceivable application:

Use the XORI (exclusive OR immediate) instruction and the immediate value -1 to perform bit-wise inversion (not operation) on register $t2:

XORI $t2, $t2, -1

The NOR instruction can be used to carry out the following operation between registers $11 and $12 and save the outcome in $t2:

NOR $t2, $11, $12

With the assumption that $t1 holds the memory address, we can use the SW (store word) instruction to store the output of NOR into memory at address $t1:

SW $t2, 0($t1)

The NORI (NOR instant) instruction can be used to execute a NOR operation between the value in $t1 and the immediate value -12 and store the outcome in $12:

NORI $12, $t1, -12

The NOR instruction can be used to execute a NOR operation between the values in $t1 and $t2 and then store the outcome back in $t1:

NOR $t1, $t1, $t2

Thus, the bare minimum set of MIPS instructions required to execute the provided pseudoinstruction is shown here.

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The minimum beta is given as 90 for which the tolerance of the transistor can be used.

We have the following values to solve this question with

gm of amplifier = 30mA/V

input resistance = 3000 ohm

We have transconductance as

\(\frac{I_{c} }{V_{T} } = 30mA/V\)

We have to consider the thermal voltage at Vt = 25mV when i tis occuring at the room temperature.

The collector bias would be solved by

Ic = 0.03Vt

= 0.03 x 0.025

= 0.75mA

Then we have to solve for the minimum value of the dc current as

β = gmγm

= 0.03 * 3000

= 90

Hence the minimum beta is given as 90

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Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Solution :

The dimensionless conduction heat rate, \($q_{ss}^*$\)

\($q_{ss}^*=\frac{q\times L_c}{K A_s(T_1-T_2)}$\)   ...........(1)

where \($L_c$\)  = characteristic length

                \($=\left(\frac{A}{4\pi} \right)^{1/2}. \sqrt{\frac{D^2}{4}}$\)

              A is surface area

          q = heat transfer

\($q=Sk(T_1-T_2)$\)   ..................(2)

where, S = conductor shape factor

Now substituting (2) in (1),

\($q_{ss}^* = \frac{Sk(T_1-T_2)L_c}{kA(T_1-T_2)}$\)

\($q_{ss}^* = \frac{S \times L_c}{A}$\)

\($q_{ss}^* = \frac{S \times D/2}{\pi D^2}$\)

\($q_{ss}^* = \frac{S \times D}{2\pi D^2}$\)   ...................(3)

For a sphere, we know S = 2πD

Putting this in (3),

\($q_{ss}^* = \frac{2 \pi D \times D}{2\pi D^2}$\)

\($q_{ss}^* = \frac{2 \pi D^2}{2\pi D^2}$\)

\($q_{ss}^* = 1$\)

Therefore, the dimensionless heat conduction rate for a sphere is 1.

Answer:

The strain of the steel rod due to the applied force is 41.93

Explanation:

Modulus of elasticity is equal to stress divided by strain.

And stress is equal to force divided by area

Surface area of cylindrical rod

\(2\pi r (r+h)\)

Substituting the given values we get -

\(2 *3.14 * \frac{1.5}{1000} * 45 (45 + \frac{1.5}{1000}) = 19.07\)

\(2 * 10 ^5 = \frac{250}{19.07 * S=(\frac{\Delta L}{L} )}\)

Hence, strain is equal to

Strain = 41.93

Answer:

The temperature of the outer wall surface will be closer to the temperature of the surrounding air.

Explanation

Since the outer wall has a convective heat transfer coefficient that is 3 time the one in the inner surface due to the wind, the outer wall will have a temperature closer to the surrounding air's temperature because the heat that flows from the warmer inner room wall will be rapidly conducted away from the surface of the outer wall by the prevailing wind, leaving it with a reduced temperature that will be closer to the temperature of the surrounding air.

Answer:

  A accountant

  D teacher

  B cook

  C beautician

Explanation:

C is on the right of the accountant, so D (opposite C) is the teacher. The accountant is to the right of the teacher, so cannot be B, and must be A. Then B (opposite A) is the cook, and C (left of B) is the beautician.

__

Clockwise from A:

  A accountant

  D teacher

  B cook

  C beautician

The discrepancy between experimental and theoretical entropy values can be used to determine residual entropy.

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We can use the formula for the nth term of an arithmetic sequence to find a10. Let d be the common difference, then a10 = a1 + 9d.

Using the given information, we can create a system of equations:
a1 + a3 + a5 = -12
a1 + (a1 + 2d) + (a1 + 4d) = -12
3a1 + 6d = -12
a1 + 2d = -4
a1 = -4 - 2d
a1a3a5 = 80
(-4-2d)(-4+2d)(-4+4d) = 80
-16 + 16d^2 - 64d + 16d^3 = 80
2d^3 + 8d^2 - 8d - 7 = 0
Using synthetic division or long division, we can factor the polynomial as:
(2d+1)(d+1)(d-7) = 0
Thus, the possible values for d are -1/2, -1, and 7.
Substituting each value of d into a1 = -4 - 2d, we get:
a1 = 3, -2, -18
Finally, we can find a10 for each possible value of a1:
a10 = 21, -16, -142; or
a10 = -2/3, -11, -145/3.
Therefore, the possible values of a10 are 21, -16, -142, -2/3, -11, and -145/3.

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Tensile stress in a thin wall spherical shell is given by

σ = ( p r ) /2t

This is the type of stress that involves pulling out. it is the opposite of compressive stress.

The tensile stress, σ  at the circumference of the shell is the hoop stress. The value equals that of pressure acting on the area.

Therefore:

σ * 2πr * t = p * πr^2

where

2πr = circumference of a circle

πr^2 = area of a circle

p = internal pressure

t = thickness

σ * 2πr * t = p * πr^2

σ = ( p r ) /2t

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a. Yes, because you will be working in a team of 10 design engineers and take note some of them are experienced and capable nuclear plant designers.And from what i have learn about engineers, they can't be called engineers for nothing.All you have to do is trust yourself and your team.

b.Of course i cannot but because i have my team (10 design engineers) i know i can do a great job and they are experienced in this stuff so i am confident that i can participate and can make a good job and even if i do not have any experience i know i will learn from them.I can call this assignment as my experience right?

Nuclear power plants uses nuclear fuels as a source of energy.

A nuclear power plant is a power generating plant that makes use of nuclear fuels as a source of energy for power generation. This means of power generation has the advantage that 1 g of nuclear fuel can be used for several years.

On the downside, this means of power generation is prone to hazards such as nuclear explosions and radioactive fallout. This means of power generation is ethical as long as all the safety standards are followed.

I can participate in such a design as a trainee engineer without any previous experience because participating in such project is a huge learning opportunity for me.

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The volumetric pipette is the most accurate device to measure the volume of a standard solution. Therefore, the correct answer is (c) i.e. volumetric pipette.

The volumetric pipette is the most accurate pipette out of all for measuring volumetric annalysis. Volumetric pipettes are glasswares that are designed to transfer a particular small volume of sample fluid from one container to another in a very stable and controlled manner while preserving an accurate record of each sample drop.

What are pipettes?

In the context of the laboratory, pipettes are used to transfer liquid from one container to another accurately and swiftly.  While there exist several different types of pipettes, it is vital to take into account that some certain pipettes offer more accuracy as a comparison to others.

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|H(omega)| = 1/(sqrt(1 + (omegaRC) ² )) is the magnitude of the frequency response, which is a standard form for a first-order low-pass filter, and the cutoff frequency is given by omega_c = 1/(R  ˣ C).

What is the frequency response of the given circuit and how is it calculated?

To determine the frequency response of the given circuit, we need to find the transfer function H(omega) = V/V_s, where V is the output voltage and V_s is the input voltage.

We can start by analyzing the circuit using the voltage divider rule. The impedance of the resistor R is simply R, while the impedance of the capacitor C is given by 1/(jˣ omegaˣ C), where j is the imaginary unit. Thus, the voltage across the capacitor can be expressed as:

V_c = V_s   ˣ  1/(1 + j  ˣ omega  ˣ R  ˣ C)

The output voltage V is simply the voltage across the capacitor, so:

V = V_c = V_s  ˣ 1/(1 + j  ˣ omega  ˣ R  ˣ C)

Now we can calculate the frequency response by taking the magnitude of H(omega):

|H(omega)| = |V/V_s| = 1/(sqrt(1 + (omega  ˣ R  ˣ C) ² ))

This is a standard form for the frequency response of a first-order low-pass filter, where the cutoff frequency is given by:

f_c = 1/(2  ˣ pi  ˣ R  ˣ C)

In terms of omega, the cutoff frequency is:

omega_c = 1/(R  ˣ C)

To sketch |H(omega)| versus omega, we can plot the function 1/(sqrt(1 + (omega  ˣ R  ˣ C) ² )) for omega > 0. The plot will look like a decreasing curve starting from 1 at omega = 0 and approaching 0 as omega goes to infinity. The cutoff frequency omega_c marks the point where the magnitude of the frequency response drops to 1/sqrt(2) or -3 dB.

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The primary majority of sUAS utilize GPS to coordinate base location and tracking in the United States, sUAS beyond visual line of sight inspections of pipelines and power lines are routinely carried out every day.

The meaning of GPS is Global Positioning System. This is a radio navigation system that is satellite-based. This system is installed on mobile phones and computers. GPS can tell the position of that phone or the person using that phone.

Thus, In the United States, sUAS beyond the optical line of sight inspections of pipelines and electrical lines are routinely carried out every day. The vast majority of sUAS use GPS to coordinate base position and tracking.

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Answer:

Subsystem denotes to the installation of some subassemblies, which can be very small or gigantic in size, that serve this subsystem and the overall system. ... In addition, availability importance measures are employed to rank various subsystems with regards to their impact on the overall system availability.

Explanation:

Subsystem refers to the installation of a few subassemblies that support this subsystem and the entire system.

An assembly is a template for a corridor's cross section at a certain station.

A set of interconnected subassemblies that are each connected to a center point or to other subassemblies makes up an assembly. An assortment of shapes, linkages, and points make up a subassembly.

The installation of a few subassemblies that support this subsystem and the overall system is referred to as a subsystem.

These subassemblies can range in size from little to huge. In addition, subsystems are ranked according to their impact on the overall system's availability using availability importance criteria.

The primary memory, the input/output subsystem, and the central processor unit (CPU) are the three basic categories or subsystems that make up a computer. Data processes are carried out by the central processing unit (CPU).

A subassembly could be:

Thus, these are important subassemblies and subsystems.

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Octahedral metal complexes have six sigma (σ) bonds between the metal center and the surrounding ligands.

Octahedral metal complexes have six sigma (σ) bonds between the metal center and the surrounding ligands. These sigma bonds are formed through the overlap of atomic orbitals between the metal and the ligands, resulting in a symmetrical, octahedral arrangement of the ligands around the central metal atom.

In addition to sigma bonds, octahedral metal complexes may also have pi (π) bonds, which are formed through the overlap of the p-orbitals of the ligands and the d-orbitals of the metal. Pi bonds are generally weaker than sigma bonds and may contribute to the electronic and optical properties of the complex.

However, in octahedral complexes, there are no sigma * (σ*) bonds. Sigma * bonds are formed by the overlap of antibonding molecular orbitals, which have a nodal plane that runs through the bonding axis, resulting in destructive interference and weakening of the bond. In octahedral complexes, there are no antibonding orbitals that overlap along the bonding axis, so there are no sigma * bonds.

Overall, octahedral metal complexes have six sigma (σ) bonds and may also have pi (π) bonds, but they do not have any sigma * (σ*) bonds.

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Where the above conditions are given for the crystal structure referenced above, the density of the crystal structure is: 5.068 x 10⁻²¹ g/nm³.

The density of a crystal structure is the mass per unit volume of its unit cell. It is important for understanding the physical and chemical properties of materials. It is often used in material science and engineering to determine the quality and suitability of a crystal for various applications.

The equation for density is:

density = (A * n) / (a³ * Na)

where:

Given that:

n = 1,

a = 2R,

A = 74.5 g/mol,

R = 0.145 nm, and

Na = 6.022 x 10²³, we can substitute the values into the equation:

Density = (74.5 * 1) / ((2R)³ * 6.022 x 10²³)

Density = (74.5 * 1) / ((2 * 0.145)³ * 6.022 x 10²³)

Density = 74.5/ (0.024389 * 6.022 * 10²³)

Density = 74.5 / (1.46957 x 10²²)

Density = 5.068 x 10⁻²¹ g/nm³

Thus, the density of the crystal structure is: 5.068 x 10⁻²¹ g/nm³.

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Full Question:

Although part of your question is missing, you might be referring to this full question:

For a simple cubic crystal structure, the value of n = 1 since there is only one atom associated each unit cell.

Furthermore, for the unit cell edge length a = 2R and we have A = 74.5 g/mol & R = 0.145nm, using equation, find the density of the structure.