The rotating magnetic field in an AC machine is produced by AC voltages applied to the stator windings. DC voltages applied to the stator windings. AC voltages applied to the rotor windings. special permanent magnets on the stator.

To find the tension in the string, we need to consider Newton's second law of motion and the force of friction acting on the block. The tension in the string is approximately 20 N (option C).

First, let's find the force of friction (F_friction) using the formula: F_friction = μ * F_normal, where μ is the coefficient of kinetic friction and F_normal is the normal force acting on the block, which is equal to the weight of the block (m * g).

F_friction = 0.20 * (4.0 kg * 9.80 m/s²) = 0.20 * 39.2 N = 7.84 N

Next, let's use Newton's second law of motion, F_net = m * a, where F_net is the net force acting on the block, m is the mass of the block, and a is the acceleration of the block.

The net force acting on the block is the difference between the tension in the string (T) and the force of friction: F_net = T - F_friction.

We know that the acceleration (a) is 3.0 m/s², and the mass (m) is 4.0 kg. So, we can rewrite the equation as:

4.0 kg * 3.0 m/s² = T - 7.84 N

12 N = T - 7.84 N

Now, solving for the tension in the string (T):

T = 12 N + 7.84 N = 19.84 N

The tension in the string is approximately 19.84 N, which is closest to option C. 20 N.

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The correct answer for the given conditions is D. If the ball is not orange, then the ball bounces.

The original statement is "If the ball does not bounce, then the ball is orange." To find the negation of this statement, we need to negate both the condition and the consequence.

The condition "the ball does not bounce" can be negated as "the ball bounces." This means that in the negation, we consider the case where the ball does bounce.

The consequence "the ball is orange" can be negated as "the ball is not orange." This means that in the negation, we consider the case where the ball is not orange.

Therefore, the negation of the original statement is "If the ball bounces, then the ball is not orange." This can be written as "If the ball is not orange, then the ball bounces."

Option D, "If the ball is not orange, then the ball bounces," correctly represents the negation of the original statement.

In option A, the negation incorrectly states that if the ball bounces, then the ball is not orange, which is the same as the original statement.

In option B, the negation states that the ball does not bounce and the ball is not orange, which is a combination of negating both the condition and the consequence.

In option C, the negation states that the ball bounces and the ball is not orange, which again combines both the original statement and its negation.

Therefore, option D is the correct negation of the original statement.

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Answer:

11.85 k

Explanation:

11.85 k

The electric field strength across the membrane is approximately 8.612 × 10⁶ V/m, given a voltage of 77.7 mV and a membrane thickness of 9.02 nm.

To determine the electric field strength across a membrane, we can use the formula:

Electric field strength = Voltage / Distance

Given that the voltage across the membrane is 77.7 mV (millivolts) and the membrane thickness is 9.02 nm (nanometers), we need to convert the units to be consistent.

1 mV = 0.001 V (volts)

1 nm = 1e-9 m (meters)

Converting the units:

Voltage = 77.7 mV × 0.001 V/mV = 0.0777 V

Distance = 9.02 nm × 1e-9 m/nm = 9.02e-9 m

Plugging the values into the formula:

Electric field strength = 0.0777 V / 9.02e-9 m

Calculating the electric field strength:

Electric field strength = 8.612 × 10⁶ V/m

Therefore, the electric field strength across the membrane is approximately 8.612 × 10⁶ V/m.

In summary, the electric field strength across the membrane is approximately 8.612 × 10⁶ V/m when the voltage across the membrane is 77.7 mV and the membrane thickness is 9.02 nm.

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Answer:

Wasted energy is energy that is not usefully transferred or transformed. Energy cannot be made or destroyed. Energy is transformed into a different form that can be used. When energy is transformed or transferred only part of it can be usefully transformed or transferred.

Answer:

atom

Explanation:

the first guy who gave link this is for you:

we cannot use links in brainly

Thankyou.

Answer:

i think its electrostatic force

The empty shopping cart is moving at 7.67 m/s.

According to the law of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Total initial momentum = total final momentum

100 kg ×5.0 m/s + 30 kg ×0 m/s = 100 kg ×2.7 m/s + 30 kg ×v

v = (5.0 - 2.7)(100/30) m/s

= 7.67 m/s.

Hence,  the empty shopping cart is moving at  speed of 7.67 m/s.

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The irrigation frequency is approximately 88.89 and the irrigation period is 3000 hours.

I. To determine the localized ETc (crop evapotranspiration) at peak demand for matured grape plantation, we can use the following formula:

ETc = ETo × Kc × Kr × Groundcover

Given:

ETo (reference evapotranspiration) at peak demand = 6.5 mm/d

Kc (crop coefficient) = 1.2

Kr (reduction coefficient) = 0.94

Groundcover = 80% (0.8)

Calculating:

ETc = 6.5 × 1.2 × 0.94 ×0.8

ETc = 5.11744 mm/d

Therefore, the localized ETc at peak demand for the matured grape plantation is approximately 5.12 mm/d.

II. To calculate the peak net and gross water requirements for mango plantation on sandy soil, we use the formula:

Net Requirement = ETc / Ks

Gross Requirement = Net Requirement / EU

ETc (localized ETc) = 5.12 mm/d

Ks (soil water stress coefficient) = 0.91

EU (water use efficiency) = 85% (0.85)

Calculating Net Requirement:

Net Requirement = 5.12 / 0.91

Net Requirement = 5.62 mm/d

Calculating Gross Requirement:

Gross Requirement = 5.62 / 0.85

Gross Requirement = 6.62 mm/d

Therefore, the peak net water requirement for mango plantation on sandy soil is approximately 5.62 mm/d, and the peak gross water requirement is approximately 6.62 mm/d.

b. To determine the number of emitters per plant, we can use the following formula:

Number of emitters = Wetted area per plant / Wetted area per emitter

Tree spacing = 6 m x 6 m

Percent wetted area (Pw) = 50%

Wetted area for sandy soil = 2 m²

Calculating Wetted area per plant:

Wetted area per plant = Tree spacing ×Tree spacing ×Pw

Wetted area per plant = 18 m²

Calculating Number of emitters per plant:

Number of emitters per plant = Wetted area per plant / Wetted area per emitter

Number of emitters per plant = 18 / 2

Number of emitters per plant = 9 emitters

Therefore, the number of emitters per plant is 9.

c. To determine the irrigation frequency and irrigation period, we need to consider the effective rooting depth, soil available moisture content, manageable allowable depletion, and drip emitter discharge.

Effective rooting depth = 1000 mm

Soil available moisture content = 15 cm/m

Manageable allowable depletion = 25% (0.25)

Drip emitter discharge = 0.005 m³/h

Calculating Irrigation Frequency:

Irrigation Frequency = Effective Rooting Depth / (Soil Available Moisture Content × (1 - Manageable Allowable Depletion))

Irrigation Frequency = 1000 / (15 × (1 - 0.25))

Irrigation Frequency = 1000 / (15 × 0.75)

Irrigation Frequency = 1000 / 11.25

Irrigation Frequency ≈ 88.89

Calculating Irrigation Period:

Irrigation Period = Wetted Depth / Drip Emitter Discharge

Irrigation Period = 15 / 0.005

Irrigation Period = 3000 hours

Therefore, the irrigation frequency is approximately 88.89 and the irrigation period is 3000 hours.\

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Answer: False

Explanation:  

Answer:

conduction s convention s radiation

The currents through and voltages across the resistors in the circuit are:

I₁ = 1.5 A, V₁ = 60 VI₂ = 3 A, V₂ = 60 VI3 = 1 A, V₃ = 30 V, V₄ = 10 V

How we can find currents and voltages across the resistors?

voltage sources e₁ = 60 V,

resistors R₁ = 40 Ω, R₂ = 20 Ω, R₃ = 30 Ω and R₄ = 10 Ω.

The given circuit is as follows:

The current I₁ flows through resistor

R1.I1

= e₁/R₁

= 60/40

= 1.5 A

The current I₂ flows through resistor R₂.I₂

= e₁/R₂

= 60/20

= 3 AI₃ flows through resistors R₃ and R₄.

I₃ = e₁/(R₃+R4)

= 60/(30+10)

= 1 AThe voltage across R₁ is given by V₁

= I₁R₁ = 1.5 × 40

= 60 V

The voltage across R₂ is given by V₂

= I₂R₂

= 3 × 20

= 60 V

The voltage across R₃ is given by V₃

= I₃R₃

= 1 × 30

= 30 V

The voltage across R₄ is given by V₄

= I₃R₄

= 1 × 10

= 10 V

Therefore, the currents through and voltages across the resistors in the given circuit are:

I₁ = 1.5 A, V₁ = 60 VI₂ = 3 A, V₂ = 60 VI3 = 1 A, V₃ = 30 V, V₄ = 10 V

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Following the image of the ball that have been shown;

1. Potential energy

2. Kinetic energy

3. Kinetic energy

4. Potential energy

5. Kinetic energy

6. Kinetic energy

7. Kinetic energy

Due to its position or elevation, an object has potential energy while it is at rest or is elevated above the ground. Depending on the circumstances, this potential energy may either be elastic or gravitational.

Kinetic energy, or the energy connected to motion, is present when an item is in motion. An object's kinetic energy is determined by its mass and velocity .

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Answer:

1: Potential and kinetic

2: Potential and kinetic

3: Potential and kinetic

4: potential

5: Potential and kinetic

6: Potential and kinetic

7: NEITHER

Explanation:

4 is potential because its the highest point

7 is neither because the question states "the image shows the path of a ball from the time it's thrown to the time it lands on the ground" so... 7 is when the ball has landed on the ground. therefore it has neither potential nor kinetic because it is resting.

Answer:

A . Force

C . Velocity .

D . Acceleration .

If the static friction coefficient is 0.20 and you are riding a motorcycle on wet surface, the quickest you could go before losing traction is 60.76 m/s.

Friction plays a critical part in maintaining the vehicle's slipping on the road.

The formula v = Rg, where v is the vertical movement, determines the maximum speed of a roadway with friction.

The static friction coefficient is.

The circle road's radius is R.

Gravitational acceleration, or g,

Assuming that = 0.20 R = 31 m g = 9.8 m/s2

When the supplied values are substituted in the calculation above, the result is = 0.20 319.8v = 60.76m/s.

Therefore, the wet road's maximum speed is 147 m/s.

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Answer:

Netball is a ball sport for two teams of seven players it's rules are published in print and online by the international netball federation Games are played on a rectangular court divided into thirds with a raised goal at coach short end

Explanation:

It will help you

Answer:

Explanation:

Velocity is at its greatest when kinetic energy is at its max which is when all the ball's energy has been transformed from potential energy to kinetic energy which is at the lowest point in its travels (assuming the ball is rolling down a ramp). You have no picture here so this answer is a general one, not a specific one.

Here work done by the car against the force of air in each second denotes the power needed to overcome air resistance at that particular speed. The power of engine is 28 kW.

The power of an object denotes the rate of performing the work. It is defined as the work done in unit time. The SI unit of power is Watt (W) which is also called joules per second.

Power = Resistance × Speed of vehicle

Here the air resistance is found to be 700 N. Then,

Power = 700 × 40 = 28000 N m/s = 28 kW

Thus the engine power is 28 kW.

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When a car rounds a curve, it experiences a centripetal acceleration towards the center of the curve. The magnitude of this acceleration depends on the speed of the car and the radius of the curve. The magnitude of the car's acceleration is  \(5.0 m/s^2\), which is option (C).

We can use the centripetal acceleration formula to solve this problem:

a =\(v^2 / r\)

where a is the centripetal acceleration, v is the speed of the car, and r is the radius of the curve.

Plugging in the given values, we get:

a =\((10 m/s)^2 / 20 m\)

a = \(5 m/s^2\)

Therefore, the magnitude of the car's acceleration is \(5.0 m/s^2\),  which is option (C).

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For a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin is the conclusion which can be made from Gay-Lussac’s law and is denoted as option D.

This is referred to as the perpendicular force applied on a body per unit area and the unit is Pascal.

Gay-Lussac was a scientist who discovered through numerous experiments and observations that at a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin.

This is denoted as the following equation below:

P ∝ T

P = kT where p is pressure, k is constant and t is temperature.

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The options are:

Explanation:

because conduction takes place in objects whose particles are close/touching each other therefore conduction happens in solid as liquid and gas paricles are not close to each other

We can observe that the ball reaches the bottom with the highest speed when the ball is dropped.

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Discovery Statement:

I discovered that my note-taking skills needed improvement when I attended a lecture by a guest speaker who spoke very quickly and used complex technical terms that were unfamiliar to me. I found it challenging to keep up with the speaker's pace and understand the content at the same time. As a result, I struggled to take accurate and organized notes, and ended up with a confusing and incomplete set of notes. When I reviewed my notes later, I realized that I had missed important points and failed to capture the essence of the lecture. This experience made me realize that I need to improve my note-taking skills, especially in situations where the content is complex and the speaker is fast-paced.

My intention is to implement the following five note-taking strategies immediately to improve my skills:

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See transcribed text below

MINDTAP

try: Transform Your Note-Taking

ms Journal Entry Transform your note taking

3

Q Search this course

0

Q

Question 1 of 1

Think back on the last few lectures you have attended. How would you rate your note-taking skills? As you complete this exercise, think of areas that need improvement.

1. Discovery Statement

First, recall a recent incident in which you had difficulty taking notes. Perhaps you were listening to an instructor who talked fast. Maybe you got confused and stopped taking notes altogether. Or perhaps you went to review your notes after class, only to find that they made no sense at all. Describe this incident in more detail, noting how it was challenging for you.

I discovered that...

X

2. Intention Statement

Now review this chapter to find at least five strategies that you can use right away to help you take better notes. In the space below, sum up each of those strategies in a few words, and note the title of the article where these strategies are explained.

2 Action Statement

Q Search

LM

A-Z

b

بین

9:19 PM

3/3/2023

Background: Nuclear energy can be used for various industrial applications, such as seawater desalination, hydrogen production, district heating or cooling, the extraction of tertiary oil resources and process heat applications such as cogeneration, coal to liquids conversion and assistance in the synthesis of chemical ...

Answer:

Biosphere

Explanation:

Plants release the biosphere into the atmostphere. :)

Answer:

Work

Explanation:

The use of force to move an object is the definition of "work" in physics.

Answer:

the answer should be wave

The distance between point A and the point charge is approximately 1.815 micrometers.

The electric potential at a point in the electric field created by a point charge is given by the formula V = kq/r, where V is the electric potential, k is the Coulomb constant (9.00 x \(10^{9}\) \(Nm^{2}/C^{2}\)), q is the point charge, and r is the distance from the point charge.

Rearranging this equation, we get r = kq/V. Plugging in the given values, we get: r = (9.00 x \(10^{9}\) \(Nm^{2}/C^{2}\))(3.3 x \(10^{-11}\) C)/(0.6 V)

Simplifying this expression, we get: r = 1.815 x \(10^{-6}\) m

Therefore, the distance between point A and the point charge is approximately 1.815 micrometers.

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Answer:

a = 5 [m/s²].

Explanation:

To solve this equation we must use the following equation of kinematics.

\(v_{f} =v_{o} +a*t\)

where:

Vf = final velocity = 25 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 4 [s]

Note: The positive sign for the acceleration in the above equation means that the car is increasing its velocity.

25 = 5 +a*(4)

25 - 5 = 4*a

20 = 4*a

a = 5 [m/s²]

If the tension in rope is 145 N, work done to move crate 4.0 m is 442.4 J.

To solve this problem, we first need to find the force acting on the crate in the horizontal direction. This force is equal to the tension in the rope multiplied by the cosine of the angle between the rope and the horizontal:

F_horizontal = Tension * cos(angle) = 145 N * cos(40.8 degrees) = 110.6 N

Next, we can calculate the work done to move the crate using the formula:

Work = Force * Distance * cos(theta)

where theta is the angle between the force and the direction of motion. Since the crate is moving with a constant speed, we know that the net force on it is zero. Therefore, the force of friction acting on the crate must be equal in magnitude to the force we calculated above:

F_friction = F_horizontal = 110.6 N

The angle between the force of friction and the direction of motion is 180 degrees, so we have:

Work = F_friction * Distance * cos(180 degrees) = - 110.6 N * 4.0 m * cos(180 degrees) = - 442.4 J

The negative sign indicates that the work done is in the opposite direction to the displacement of the crate. In other words, the work done by the force of friction is negative because it acts against the motion of the crate. Therefore, the work done to move the crate 4.0 m is -442.4 J.

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Answer: A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s2. A spaceship far from any star or planet accelerates uniformly from 72 m/s to 160 m/s .

Explanation: i hoped that helped you.