A 2-bit positive-edge triggered register has data inputs d1, d0, clock input clk, and outputs q1, q0. Data inputs d1d0 are 01 and outputs q1q0 are 00. What do q1q0 become after the rising edge of the clock occurs

Answer:

Exploration geophysics is an applied branch of geophysics and economic geology, which uses physical methods, such as seismic, gravitational, magnetic, electrical and electromagnetic at the surface of the Earth to measure the physical properties of the subsurface, along with the anomalies in those properties

The FCC unit cell is a cube with lattice points at each corner. The points can be defined by: (0, 0, 0), (0.5, 0.5, 0), (0.5, 0, 0.5), (0, 0.5, 0.5), (0, 0, 0.5), (0.5, 0, 0), (0, 0.5, 0), (0.5, 0.5, 0.5).

What is FCC unit cell?
FCC unit cell is a three-dimensional representation of a crystal lattice, consisting of a unit cell which is repeated in all directions. It is one of the most common types of unit cells and is also known as a cubic unit cell. The unit cell is a cube with atoms located at the corners and at the center of each face. These atoms form a lattice pattern which is repeated in all directions. The FCC unit cell has a lattice constant, or unit cell edge length, of a=4R, where R is the atomic radius of the atoms in the lattice. The FCC unit cell has a total of four lattice points per unit cell, which is why it is known as a body-centered cubic (BCC) unit cell. The FCC unit cell is widely used in materials science to describe the structure of many metallic and non-metallic materials.

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There are a lot of ways employees uses in controlling exposure routes.  But when risk assessment is not be performed is not a part of the control methods.

The ways to control workplace hazards are known to be means taken to ensure safety in the workplace.

The examples are:

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Answer:

which class is this. I don't know sry

In a scenario where you have been working as a network engineer for Flaming Systems and Solutions. An existing client of your organization has requested you to set up a network for a new joint wherein a user will have to agree to a certain set of terms and conditions before gaining access to the network, technology you should use in such a scenario is: Captive portal.

A captive portal is a web page that is viewed via a web browser and presented to newly joined Wi-Fi or wired network users before they are permitted broader access to network resources.

Captive portals are the welcome pages that appear when you join to a Wi-Fi hotspot in a public location such as an airport or your favorite fast-food restaurant.

The basic line is that when a person is utilizing a captive gateway, it doesn't take long for a hacker to obtain access to their device. Once attackers have access to that device, it can serve as a back door to the actual prize, which is your business network, as well as the networks of your partners and/or supplies chain.

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Answer: true

Explanation: I haves

When MIPS detects an overflow, it raises an unscheduled procedure call, known as an interrupt or an exception, to the Operating system.

An interrupt is a signal that is sent to the processor by a device, program, or other process, indicating that an event needs immediate attention. When an overflow occurs, MIPS may raise an interrupt to alert the operating system of the issue.

An exception is a type of interrupt that occurs when the processor detects an error or an unusual condition in the current instruction or program. In the case of an overflow, an exception would be raised to notify the operating system that the calculation cannot be completed as expected.

The operating system can then handle the exception accordingly, for example, by terminating the program or taking corrective action.

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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

1. Given the below input signal, VIN(t), and the phasors of the transfer function at various frequencies, calculate the expression for the output signal, VOUT(t) in the form VOUT(t)=Acos(200πt+θ1)+Bcos(400πt+θ2).

What is the value of A?

VIN(t)=10cos(200πt+30)+20cos(400πt+45)

H(f)=0.1∠10      f=100Hz

H(f)=1∠−20      f=200Hz

H(f)=2∠30      f=300Hz

H(f)=3∠−40     f=400Hz

H(f)=4∠50      f=500Hz

2. In the same question, what is the value of θ1 in degrees? Enter the value in the box below without the units.

3. In the same question, what is the value of B? Enter the value in the box below without the units.

4. In the same question, what is the value of θ2 in degrees? Enter the value in the box below without the units.

Answer:

Vout(t) = 10cos(200πt + 10°) + 60cos(400πt + 5°)

1. What is the value of A?

A = 10

2. In the same question, what is the value of θ1 in degrees?

θ₁ = 10°

3. In the same question, what is the value of B?

B = 60

4. In the same question, what is the value of θ2 in degrees?

θ₂ = 5°

Explanation:

The output signal Vout(t) is in the form

Vout(t)=Acos(200πt + θ₁) + Bcos(400πt + θ₂)

The input signal Vin(t) is given as

Vin(t) = 10cos(200πt + 30°) + 20cos(400πt + 45°)

The output signal Vout(t) is found by

Vout(t) = H(f) × Vin(t)

Where H(f) is the transfer function at various frequencies and Vin(t) is the input signal.

H(200) = 1 ∠−20°

H(400) = 3 ∠−40°

converting the input signal into phasors

10cos(200πt + 30°) = 10 < 30°

20cos(400πt + 45°) = 20 < 45°

Vout(t) = H(200)×10cos(200πt + 30°) + H(400)×20cos(400πt + 45°)

Vout(t) = (1 ∠−20°)×10cos(200πt + 30°) + (3 ∠−40°)×20cos(400πt + 45°)

Vout(t) = (1 ∠−20°)×(10 < 30°) + (3 ∠−40°)×(20 < 45°)

Multipy the magnitude and add phase angles together

Vout(t) = (1×10 ∠−20° + 30°) + (3×20 ∠−40° + 45°)

Vout(t) = 10 ∠ 10° + 60 ∠5°

Vout(t) = 10cos(200πt + 10°) + 60cos(400πt + 5°)

Comparing it with the general form

Vout(t)=Acos(200πt + θ₁) + Bcos(400πt + θ₂)

1. What is the value of A?

A = 10

2. In the same question, what is the value of θ1 in degrees?

θ₁ = 10°

3. In the same question, what is the value of B?

B = 60

4. In the same question, what is the value of θ2 in degrees?

θ₂ = 5°

Answer:

they were updated rather than being created

Answer:

Invention is about creating something new, while innovation introduces the concept of “use” of an idea or method.

Answer:

142.825 m³/sec

Explanation:

Let the discharge of this stream = Q1

200g/L = 200x10³mg/L

Q2 = 200x10³ ppm

C1 = 10

C2 = 0.025

Qmix * Cmix = c1Q1 + c2Q2

(Q1+0.025)45 = 10xQ1 + 0.025*200x10³

45Q1 + 1.125 = 10Q1 + 5000

45Q1 -10Q1 = 5000 - 1.125

35q1 = 4998.875

Q1 = 4998.875/35

Q2 = 142.825 m³/sec

142.825 m³/sec is therefore the estimated discharge from the stream.

there is no smoke its a electric train

Answer: They are all styles of reggae drumming.

In an n type semiconductor, the Fer mi level is located just below the conduction band in the band gap.

In an n type semiconductor, the Fe rmi level refers to the energy level within the material that represents the highest probability of finding an electron. The Fer mi level plays a crucial role in determining the electrical behavior of the semiconductor.

In an n type semiconductor, the material is made with impurities that introduce additional electrons into the crystal lat tice, creating an excess of negatively charged carriers (electrons).

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No, generally it is not. It could be zero for specific loading conditions. Since they are equal but with opposite signs then their average is zero.

For example, if you put a stress element under tension in the x direction and compress it in the y direction with equal stresses such as sigma_x=sigma_y, this will give you a stress element if rotated by 45 degrees will give you the maximum shear planes with zero normal stresses. So generally that's not the case. The general case is that you will get the average normal stress when you get the maximum shear stress. In the example I mentioned before if we consider the compression and tension natures of the stresses, we assumed then sigma_y should be negative because it's compression and sigma_x is positive because it's tension.

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For fuel gas and oil applications, the maximum operating pressure with MegaPress G is 125 psi (pounds per square inch). MegaPress G is specifically designed for use with gas and oil applications, providing a safe and efficient method of connecting pipes while ensuring proper pressure management.

Maximum operating pressure with MegaPress G for fuel gas or oil applications: 125 psi (pounds per square inch) is the maximum operating pressure with MegaPress G for fuel gas and oil applications. This indicates that MegaPress G is made to safely withstand pressures of up to 125 psi in applications involving gasoline, gas, and oil.

MegaPress G is specifically intended for use in gas and oil applications. It is therefore tailored to the special qualities of these materials and built to withstand the strains and pressures that are typical in these applications.

Provides a method for safely and effectively joining pipes: For joining pipes in gas and oil applications, MegaPress G offers a secure and effective solution. The press fitting approach does not require threading, soldering, or welding, which can be labour-intensive and even dangerous. Instead, a secure and dependable connection is made by compressing the fittings onto the pipe using a press tool.

Assures correct pressure management: In gas and oil applications, where high pressures may be present, proper pressure management is crucial. MegaPress G is made to provide a safe and dependable connection between pipes to maintain optimum pressure management. Leakage, which can result in pressure decreases and other problems are reduced as a result.

In conclusion, MegaPress G was created primarily for use in gas and oil applications, and it is capable of handling pressures of up to 125 psi without harm. It offers a reliable and safe way to join pipes while guaranteeing correct pressure control to reduce the possibility of leaks and other problems.

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Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V\(_s\)/( 1 - D )

given that; V\(_s\) = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

\(I_L\) = V\(_s\) / ( 1 - D )²R

given that R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6

we substitute

\(I_L\) = 20 / (( 1 - 0.6 )² × 12.5)

\(I_L\) = 20 / (( 0.4)² × 12.5)

\(I_L\) = 20 / ( 0.16 × 12.5 )

\(I_L\) = 20 / 2

\(I_L\) = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

\(I_{Lmax\) = [V\(_s\) / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

\(I_{Lmax\) = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

\(I_{Lmax\) = [20 / 2 ] + [ 60 / 20 ]    

\(I_{Lmax\) = 10 + 3

\(I_{Lmax\) = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

\(I_{Lmax\) = [V\(_s\) / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

\(I_{Lmin\) = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

\(I_{Lmin\) = [20 / 2 ] -[ 60 / 20 ]    

\(I_{Lmin\) = 10 - 3

\(I_{Lmin\)  = 7 A

Therefore, the maximum inductor current is 7 A

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

\(I_D\) = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

\(I_D\) = 50 / 12.5

\(I_D\) = 4 A

Therefore, the average current in the diode under ideal components is 4 A

Answer:

\(94.25\)

Explanation:

Given: Handle of jackscrew is 30 cm long. Also, the car raised 2.0 cm with every full turn of the handle.

To find: ideal  mechanical advantage of the jack

Solution:

Let \(d_e\) denotes the distance over which the effort is applied and \(d_r\) denotes the distance that the load travels.

Here,

\(d_r=2\,\,cm\)

\(d_e=2\pi(length\,\,of\,\,the\,\,handle)=2\pi(30)=60\pi\)

So,

Ideal  mechanical advantage of the jack\(\frac{d_e}{d_r}=\frac{60\pi}{2}=30\pi=30(\frac{22}{7})=94.25\)

Die-casting is a manufacturing process that involves injecting molten metal into a mold cavity under high pressure.

The metal solidifies quickly, taking the shape of the mold, and is then ejected. This process is used to produce a variety of components with intricate shapes and high dimensional accuracy. The strength-to-weight ratio is a crucial factor in the design of such components, as it determines the load-carrying capacity of the part per unit of weight.

The strength-to-weight ratio of a die-cast part increases with decreasing wall thickness because thinner walls require less material, resulting in a lower weight. This reduction in weight is significant because the mechanical properties of a material, such as its tensile strength and elastic modulus, are typically constant regardless of its thickness. Therefore, a thinner die-cast part has a higher strength-to-weight ratio than a thicker one made of the same material.

Additionally, thinner walls increase the surface area-to-volume ratio of the part, which can improve its heat dissipation and reduce its susceptibility to thermal deformation. This is because a higher surface area allows for more efficient transfer of heat from the part to its surroundings. Moreover, thinner walls can also improve the part's ability to withstand dynamic loads, such as vibrations or impacts, due to the reduced inertia and increased damping capacity.

However, decreasing wall thickness beyond a certain point may lead to a reduction in the part's strength. This is because thinner walls are more prone to buckling and deformation under load, and may fail due to stress concentrations or material defects. Therefore, the optimal wall thickness for a die-cast part is determined by a trade-off between its strength-to-weight ratio and its structural integrity.

In summary, the strength-to-weight ratio of a die-cast part increases with decreasing wall thickness due to the reduced weight and improved heat dissipation and dynamic performance. However, the optimal wall thickness should be carefully chosen to balance these benefits with the part's structural integrity and manufacturing feasibility.

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From what my research concluded, true

Za answa iz:

True

Twust meh

Data and insight is the component of the enterprise platform will help the company capture customer information.

These are the data and knowledge gotten from analysis of customer's information in a business.

This enterprise platform is very important as it helps the organization to improve their services

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A. friendly

Nuclear engineer Meena Mutyala argues that nuclear power is an environmentally friendly technology, operating with essentially no emissions.

Nuclear is the process of splitting atoms to release energy. This energy is produced when the nucleus of an atom splits, either naturally or through a man-made process. During this process, a significant amount of energy is released as heat, light and radiation. Nuclear energy has been used in various ways, including electricity generation, medical treatments, research and industrial applications.

Nuclear energy is considered a clean, renewable source of energy, as it does not produce any greenhouse gases when it is used. However, it does come with some risks, including the potential for radiation leaks and the development of nuclear weapons. Additionally, nuclear waste must be disposed of safely, which can be difficult and expensive.

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The maximum Speed of the Sprinter from the velocity time graph of his motion is; 11.98 m/s

We are given;

Initial Speed; u = 2.5 s

Total distance; d = 100 m

Total time; T = 9.6 s

The total distance is;

d = ¹/₂(9.6 + (9.6 - 2.5) * v

where v is maximum speed.

Thus;

¹/₂(9.6 + (9.6 - 2.5) * v = 100

16.7v = 200

v = 200/16.7

v = 11.98 m/s

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Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = \(\frac{weight}{area}\)

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = \(\frac{5000}{4.45}\)

                        = 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

The natural model that scientists can study to understand the impacts of sulfur aerosol injection on the climate according to Jon Lawhead (in his presentation Geoengineering: "Climate Change by Design") is: The eruption of Mt. Pinatubo.

Geoengineering is defined as the deliberate modification of Earth's climate using science, engineering, or technology. It's also known as climate engineering or climate intervention. The goal of geoengineering is to minimize or even reverse the effects of climate change on Earth, such as warming temperatures, melting glaciers, and rising sea levels.Sulfur aerosol injection in climate engineering is a type of geoengineering that involves injecting sulfur particles into the stratosphere to reflect sunlight back into space, lowering the Earth's surface temperature.

How to understand the impacts of sulfur aerosol injection on the climate. The eruption of Mt. Pinatubo, a volcano in the Philippines, can be studied to understand the impacts of sulfur aerosol injection on the climate. In 1991, Mt. Pinatubo erupted, sending millions of tons of sulfur dioxide into the atmosphere.

The sulfur dioxide reacted with water to form sulfuric acid aerosols, which remained in the stratosphere for up to two years. The aerosols reflected sunlight, reducing the amount of sunlight reaching the Earth's surface and cooling the planet by about 0.5°C.The cooling effect of the Mt. Pinatubo eruption provides a natural experiment for studying the effects of sulfur aerosol injection on the climate. By studying the eruption's impact, scientists can learn how sulfur particles in the stratosphere influence Earth's climate and design geoengineering solutions to mitigate the effects of climate change.

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Answer:open circuit

short circuit

voc

Explanation:

Answer:

The view factor for radiation emitted by surface 2 to surface 1 is 0.1

Explanation:

Given

\(F_{12} = 0.4\)

\(A_1 = 0.01m^2\)

\(A_2 = 0.04m^2\)

Required

Determine \(F_{21}\)

To do this, we make use of the following equivalent ratio

\(A_1 * F_{12} = A_2 * F_{21}\)

Make \(F_{21\) the subject

\(F_{21} = \frac{A_1 * F_{12}}{ A_2}\)

Substitute values into the equation

\(F_{21} = \frac{0.01m^2 * 0.4}{0.04m^2}\)

\(F_{21} = \frac{0.01 * 0.4}{0.04}\)

\(F_{21} = \frac{0.004}{0.04}\)

\(F_{21} = 0.1\)