How do we select only the records with no NULL values in the "Address" column?

Answer:

Composite panel garage doors

Explanation:

Engineered lumber should not be used for composite panel garage doors. Hence, option C is correct.

Horizontal panels are connected to form a single garage door in panel lift doors, sometimes referred to as sectional doors. These panels have wheels attached to them that are positioned in tracks on either side and guide the door through a reasonably sharp turn to reposition it horizontally above the garage door opening.

Yes, however it's crucial to have a pro take their place for the finest outcomes. It can be unsafe for anybody other than a professional to install garage doors due to their complexity and difficulty.

One of the most affordable solutions on the market is an aluminum garage door. They are available in a variety of design options that can complement the exterior style of your home. Low-cost aluminum garage doors.

Thus, option C is correct.

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Answer:

stagflation

Explanation:

it was used in the article

A word which was first used during the 1970s economic crisis is stagflation.

In the 1970s, an energy crisis took place in the United States of America due to the oil embargo that was imposed on it by OPEC. This oil embargo was imposed on the United States of America by the Organization of Petroleum Exporting Countries (OPEC) in 1973 because of its role in the Arab-Israeli War.

Consequently, the economy of the United States of America experienced stagflation in the following ways:

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Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

Answer:

Lack of oxygen

Decompression sickness

temperature variation

lack of gravity

Cosmic radiations hazards

motion sickness

Explanation:

When an astronaut travels in the space he is aware of the challenges he might face during his journey. There are various test and rehearsals of travelling before a final travel takes place. An astronaut goes through many challenges which includes lack of oxygen supply, decompression and motion sickness. There is no gravity in space so an astronaut will have to be aware of the difficulties he might face during his travel. An hour on earth is 7 years long in space, so an astronaut should have patience and be able to deal with time variation.

Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, \(Q_{in}\) = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

\(\dfrac{P_1\cdot V_1}{T_1} = \dfrac{P_2\cdot V_2}{T_2}\)

\(P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2} = \dfrac{V_1}{V_2} \cdot \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot \dfrac{P_1 \cdot T_2}{T_1 }\)

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

\(Q_{in}\) = m·Cv·(T₃ - T₂)

∴ \(Q_{in}\) = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, \(\eta_{Otto}\), is given as follows;

\(\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}\)

Therefore, we have;

\(\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647\)

The thermal efficiency, \(\eta_{Otto}\) ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

\(MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}\)

Therefore, we get;

\(MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974\)

The mean effective pressure, MEP ≈ 20.974 bar

(b) Four types of internal combustion engine includes;

1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

Answer:

i need points 425677

Explanation:

yurrrrrr  awnser C

Answer:

Cool I think u should do Marvel first

Answer:

The best recommendation to prevent emissions systems failure is to regularly maintain the fuel and emissions system.

Explanation:

Explanation:

There are 8.35 pounds in a gallon of water. Water weighs 1 gram per cubic centimeter or 1 000 kilogram per cubic meter, i.e. density of water is equal to 1 000 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure.

For a kinematics link with a constant angular velocity of 1 rad/s, the coordinate of the link can be determined using the equation θ = ωt, where θ represents the angular displacement and t is the time.

The speed vector can be found by taking the derivative of the coordinate equation with respect to time. Similarly, the acceleration vector can be obtained by taking the derivative of the speed vector equation.

In this scenario, we have a kinematics link with a constant angular velocity of 1 rad/s. The coordinate of the link can be determined using the equation θ = ωt, where θ represents the angular displacement and t is the time. Since the angular velocity is constant, the coordinate of the link increases linearly with time. If we know the initial position of the link, we can calculate the coordinate at any given time.

To find the speed vector, we need to take the derivative of the coordinate equation with respect to time. Since the angular velocity is constant, the derivative of θ with respect to t is simply ω. Therefore, the speed vector of the link is constant and its magnitude is equal to the angular velocity, which in this case is 1 rad/s.

To determine the acceleration vector at a specific point, let's say point P, we need to take the derivative of the speed vector equation. Since the angular velocity is constant, the derivative of the speed vector is zero. Thus, the acceleration vector at point P is zero. This means that the link is moving at a constant speed and there is no change in its velocity at point P.

In summary, for a kinematics link with a constant angular velocity of 1 rad/s, the coordinate increases linearly with time according to θ = ωt. The speed vector is constant and equal to the angular velocity, while the acceleration vector is zero at any point on the link.

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a. The position of the ball, magnitude and direction of it's velocity at 2s are x = 44.4m , y = 39.6m, v = 24.4m/s and direction is 33.4°.

b. The height of the ball is 38.1m

c. The horizontal range R is 120.2m

Using the formulas of projectile motion, we can calculate;

(a) The position of the ball at time t = 2s is given by the following equations:

\($$x = v_0 \cos(\theta) t = 37.0 \cos(53.1^\circ) \cdot 2.00 = 44.4\text{ m}$$\)

\($$y = v_0 \sin(\theta) t - \frac{1}{2} g t^2 = 37.0 \sin(53.1^\circ) \cdot 2.00 - \frac{1}{2} \cdot 9.80 \cdot 2.00^2 = 39.6\text{ m}$$\)

The magnitude of the velocity of the ball at time t = 2s is given by the following equation:

\($$v = \sqrt{v_x^2 + v_y^2}$$\)

where:

Vx = \(v_0 \cos(\theta)$\)

Vy = \(v_0 \sin(\theta) - gt$\)

Substituting the given values into the equation, we get:

\($$v = \sqrt{(37.0 \cos(53.1^\circ))^2 + (37.0 \sin(53.1^\circ) - 9.80 \cdot 2.00)^2} = 24.4\text{ m/s}$$\)

The direction of the velocity of the ball at time t = 2s is given by the following equation:

\($$\theta = \arctan\left(\frac{v_y}{v_x}\right)$$\)

Substituting the given values into the equation, we get:

\($$\theta = \arctan\left(\frac{37.0 \sin(53.1^\circ) - 9.80 \cdot 2.00}{37.0 \cos(53.1^\circ)}\right) = 33.4^\circ$$\)

(b) The time when the ball reaches the highest point of its flight is given by the following equation:

\($$t_h = \frac{v_0 \sin(\theta)}{g} = \frac{37.0 \sin(53.1^\circ)}{9.80} = 3.81\text{ s}$$\)

The height of the ball at this point is given by the following equation:

\($$h_h = v_0 \sin(\theta) t_h - \frac{1}{2} g t_h^2 = 37.0 \sin(53.1^\circ) \cdot 3.81 - \frac{1}{2} \cdot 9.80 \cdot 3.81^2 = 38.1\text{ m}$$\)

(c) The horizontal range of the ball is given by the following equation:

\($$R = v_0 \cos(\theta) t_f = 37.0 \cos(53.1^\circ) \cdot 2 \cdot 3.81 = 120.2\text{ m}$$\)

where:

Substituting the given values into the equation, we get:

\($$t_f = \frac{2 v_0 \sin(\theta)}{g} = \frac{2 \cdot 37.0 \sin(53.1^\circ)}{9.80} = 7.62\text{ s}$$\)

Therefore, the horizontal range of the ball is 120.2m

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Answer:

they could add a play structure, with the stream they can put ducks and fish in it and picnic places

brainliest plz

Explanation:

Answer:

just took the quiz (k12) answer is...

Ask questions to identify a problem, develop a model, and carry out the plan/desgn.

Explanation:

The Vickers hardness test is a versatile technique that can be used to assess both macro and micro hardness. It has a wide load range, is appropriate for a number of applications, and is made of various materials.

1. Lay out your sample on the platform.

2. Click and drag the mouse to position the sample.

3. Scroll to bring focus.

4. Choose the test approach and load.

5. Pick a goal and a title for the position.

6. To position the indenter, use the overview camera.

7. Conclude the exam.

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The minimum pressure at which the water-supplying pump must operate is approximately 2.45 x 10⁶ Pa or 2450 bar.

To calculate the minimum pressure at which the water-supplying pump must operate, we need to use the formula for the volumetric flow rate of a fluid:

Q = A * v

where,

Q =the flow rate

A = the cross-sectional area of the hole

v = the velocity of the fluid.

We can use this formula to determine the velocity of the water, and then use the Bernoulli equation to determine the minimum pressure.

From the question:

ṁ=0.05 kg/s

ρ= 996 kg/m

d= 0.02 cm

Calculating the area of the hole:

A = (π/4) * (d²)

where,

d is the diameter of the hole, and pi is approximately 3.14.

d = 0.02 cm = 0.0002 m

A = (3.14/4) * (0.0002 m)² = 1.57 x 10⁻⁷ m²

Using the mass flow rate to calculate the velocity of water:

Q = ṁ / ρ

where,

Q = the volumetric flow rate

ṁ = the mass flow rate

ρ = the density of water.

Q = (0.05 kg/s) / (996 kg/m³)

Q = 5 x 10⁻⁵ m³/s

Using the area of the hole to find the velocity of water

v = Q / A

v = (5 x 10⁻⁵ m³/s) / (1.57 x 10⁻⁷m²)

v = 318.8 m/s

Using the Bernoulli equation to find the pressure:

P = Patm + (1/2) * ρ * v²

where,

P = the pressure

Patm = the atmospheric pressure

ρ =the density of water

v = the velocity of the water.

P = Patm + (1/2) * (996 kg/m³) * (318.8 m/s)²

Hence, the minimum pressure at which the water-supplying pump must operate is approximately 2.45 x 10⁶ Pa or 2450 bar.

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Answer:

\( T_A_C = 6.296 kN \)

\( R = 10.06 kN \)

Explanation:

Given:

\( T_A_B = 8.7 kN\)

Required:

Find the tension TAC and magnitude R of this downward force.

First calculate \( \alpha, \beta, \gamma \)

\( \alpha = tan^-^1 =\frac{40}{50} = 38. 36 \)

\( \beta = tan^-^1 =\frac{50}{30} = 59.04 \)

\( \gamma = 180 - 38.36 - 59.04 = 82.6 \)

To Find tension in AC and magnitude R, use sine rule.

\( \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} \)

Substitute values:

\(\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}\)

Solve for T_A_C:

\( T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = \)

\( T_A_C = 8.7 * 0.724 = 6.296 kN \)

Solve for R.

\( R = 8.7 * \frac{sin 82.6}{sin 59.04} = \)

\( R = 8.7 * 1.156 \)

R = 10.06 kN

Tension AC = 6.296kN

Magnitude,R = 10.06 kN

To demonstrate that a problem is NP-complete, we appear that it is within the NP course and can be decreased to a known NP-complete problem. From this deduction, all the problems are NP-complete.

To demonstrate that a problem is NP-complete, we ought to appear that it is both within the NP complexity course which is as difficult as an existing NP-complete problem. Let's analyze each issue and appear how they can be seen as generalizations of known NP-complete problems:

(a) SUBGRAPH ISOMORPHISM: This problem generalizes the SUBSET-SUM problem, where rather than subsets and wholes, we have charts and subgraphs. SUBSET-SUM could be a well-known NP-complete issue, so SUBGRAPH ISOMORPHISM acquires its NP-completeness.

(b) LONGEST Way: This problem generalizes the HAMILTONIAN Way problem, where we are inquired to find a straightforward way of length rise to the number of vertices within the chart. HAMILTONIAN Way may be a known NP-complete issue, so LONGEST Way is additionally NP-complete.

(c) MAX SAT: This problem is as of now a known NP-complete problem, so no advance verification is required.

(d) Thick SUBGRAPH: This problem generalizes the CLIQUE problem, where we are inquired to discover a total subgraph of a certain size. CLIQUE could be a well-known NP-complete problem, so Thick SUBGRAPH acquires its NP-completeness.

(e) Inadequate SUBGRAPH: This problem can be seen as the complement of the Thick SUBGRAPH issue. Since Thick SUBGRAPH is NP-complete, its complement, Scanty SUBGRAPH, is additionally NP-complete.

(f) SET COVER: SET COVER may be a known NP-complete problem, so no encouraged confirmation is required.

(g) Solid Organize: This issue can be seen as a generalization of the HAMILTONIAN CYCLE  problem. In case all did values are limited to 1 or 2, rij values are all 2, and b is equal to n, the issue gets to be identical to finding a Hamiltonian cycle. Since HAMILTONIAN CYCLE is NP-complete, the Solid Arrange issue acquires its NP-completeness.

By finding the relationship between these problems and known NP-complete problems, we will conclude that they are all NP-complete.

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Option 5 is correct. Drawn or extruded steel or aluminum pipes. Aluminum is driven through the die once it has reached a soft, pliable state.

Temperature determines whether aluminum is extruded or drawn. As we previously mentioned, cold drawing, which is carried out at normal temperature rather than a very high temperature, is applied in drawn. While extrusion simply forces metal through a die that fits the required shape, draw molding necessitates each pass to gradually reduce in size until you reach your target. This is not the case with extruded tubes and beams, which can be formed in one pass.

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The general solution of the given differential equation is  \(C_1 + C_2e^{(3x)\) + (1/6)e³x + 4x.

To find the general solution of the given differential equation, we can first solve the associated homogeneous equation, which is y" - 3y' = 0.

The characteristic equation for the homogeneous equation is obtained by assuming a solution of the form \(y = e^{(rx)\), where r is a constant. Substituting this into the characteristic equation, we get:

\(r^2 - 3r = 0\)

Factoring out r, we have:

r(r - 3) = 0

So, the solutions to the homogeneous equation are r = 0 and r = 3.

Therefore, the general solution to the homogeneous equation is given by:

\(y_h = C_1e^{(0x)} + C_2e^{(3x)\)

    = C1 + C2e^(3x)

To find a particular solution to the non-homogeneous equation, we can use the method of undetermined coefficients. Since the non-homogeneous term is e³x – 12x, we assume a particular solution of the form \(y_p\) = Ae³x + Bx + C.

Plugging this particular solution into the original differential equation, we get:

(9Ae³x + B - 3Ae³x - 3B) - 3(Ae³x + Bx + C) = e³x – 12x

Simplifying, we have:

6Ae³x - 3B - 3Bx - 3C = e³x – 12x

Equating the coefficients of like terms on both sides, we get:

6A = 1 (from the coefficient of e³x)

-3B = -12 (from the coefficient of x)

-3C = 0 (from the constant term)

Solving these equations, we find A = 1/6, B = 4, and C = 0.

Therefore, a particular solution to the non-homogeneous equation is:

\(y_p\) = (1/6)e³x + 4x

The general solution to the given differential equation is the sum of the homogeneous and particular solutions:

y = \(y_h + y_p\)

  = \(C_1 + C_2e^{(3x)\) + (1/6)e³x + 4x

This is the general solution of the given differential equation.

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Straightened or channelized stream channels can cause several problems, including Increased water velocity and Increased water velocity

Increased water velocity: Straightening or channelizing a stream removes natural meanders and bends, resulting in a straight, uniform channel. This alteration increases water velocity, which can lead to erosion and downstream flooding. The higher velocity can also impact aquatic habitats by damaging vegetation and disrupting the natural flow dynamics.

Loss of habitat diversity: Natural streams have a variety of habitat features such as pools, riffles, and meanders that support diverse ecosystems. Straightening or channelizing a stream removes these features, resulting in a loss of habitat diversity. This can negatively impact fish and other aquatic species that rely on specific habitat conditions for feeding, breeding, and shelter.

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Answer: the attached picture is the answer.

Explanation:

Assuming:

the switch position connect to 1, hence 4.5V exist at across lamp1

the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2

the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.

❎❎❎❎❎❎❎ sorry but that didn't help me that much

Electricity is generated from a nuclear fission reaction by harnessing the heat released from the splitting of atomic nuclei. In a nuclear reactor, the key components include the fuel rods, control rods, and coolant.

The fuel rods contain pellets of uranium, which undergo fission when bombarded with neutrons. The energy released from this fission heats up the coolant, typically water, which then produces steam. The steam is directed towards turbines, which drive generators to produce electricity.

The control rods are used to regulate the rate of fission reactions and prevent overheating. Overall, the process of nuclear fission reactions in a reactor generates heat, which is then converted into electricity through a series of mechanisms.

These components work together to regulate the nuclear reaction, transfer the heat produced by the reaction, and prevent the release of radioactive materials.

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The floating-point representation of 50 using the IEEE 754 Floating-Point Standard is:

0 10000100 10010000000000000000000 (or 42 90 00 00 in hexadecimal).

To convert the decimal number 50 to a floating-point number using the IEEE 754 Floating-Point Standard, we follow these steps:

Convert the decimal number 50 to binary:

50 decimal = 110010 binary

Normalize the binary representation:

110010 = 1.10010 x 2^5

Determine the sign bit:

Since 50 is positive, the sign bit is 0 (for positive numbers).

Determine the biased exponent:

The biased exponent is calculated as the exponent value plus the bias.

In IEEE 754 single-precision format, the bias is 127.

Exponent = 5, Biased Exponent = Exponent + Bias = 5 + 127 = 132

The biased exponent in binary is 10000100.

Determine the significand or mantissa:

The significand is the fractional part of the normalized binary representation, which is 1.10010.

It is written in binary as 10010 (without the leading 1).

Combine the sign bit, biased exponent, and significand:

The binary representation of 50 in IEEE 754 single-precision format would be:

0 10000100 10010000000000000000000

Breaking it down:

The first bit is the sign bit (0 for positive numbers).

The next 8 bits are the biased exponent (10000100).

The remaining 23 bits are the significand (10010000000000000000000).

Convert the binary representation to hexadecimal (optional):

The binary representation 01000010010010000000000000000000 can be converted to its hexadecimal equivalent, which is 42 90 00 00 in hexadecimal notation.

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When the membrane is depolarized, voltage-gated potassium channels do not open instantly; instead, they require roughly 1 msec to open.

The opening of a group of voltage-gated potassium channels lets potassium stream out of the cell via its electrochemical gradient. These occurrences swiftly reduce the membrane potential, returning it to its typical resting state.

Depolarization triggers the activation of voltage-gated potassium channels, and the outward flow of potassium ions through them repolarizes the membrane potential to end action potentials, hyperpolarizes the membrane potential immediately after action potentials, and is essential in establishing the resting membrane potential.

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Answer:

Mass flow rate = 15.96kg/s

Explanation:

From

P1V1=mRT1

V1=RT1/P1

V1=0.0492m³/kg

M=A(U)/V1

Where u is the velocity

A=πd²/4

M=π(0.2²)/4×25/0.0492

M=15.96kg/s

Answer:

Q ≅ 1.53 m³/s

Explanation:

From the given information:

The flow rate of the orifice is:

\(v = c_v \sqrt{2gh}\)

\(v = 0.90 \times \sqrt{2*9.81 * 10}\)

where;

\(Q = c_d \times \sqrt{2gh} \times A\); &

\(c_d = c_c \times c_v\)

∴

\(Q = c_c \times c_v \sqrt{2gh} \times \dfrac{\pi}{4}\times d^2\)

\(Q = 0.90 \times 0.62 \sqrt{2*9.81*10} \times \dfrac{\pi}{4}\times 0.5^2\)

\(Q = 0.558 \times 14.00714104 \times 0.1963495408\)

Q ≅ 1.53 m³/s

Construction methods for railway projects involve foundation work, steelwork installation, plant setup, wiring, insulation, earthing, and bonding.

The construction process for railway projects involves various steps.

By following these construction methods, including the use of high-output equipment and specialized on-track plants/machines, railway projects can be efficiently executed with a focus on safety, durability, and functionality.

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