A rescue plane flies horizontally at a constant speed searching for a disabled boat. When the plane is directly above the boat, the boat's crew blows a loud horn. By the time the plane's sound detector receives the horn's sound, the plane has traveled a distance equal to one-fourth its altitude above the ocean. Assuming it takes the sound 2.02 s to reach the plane, and taking the speed of sound to be 343 m/s, determine the following.(a) The speed of the plane. (Take the speed of sound to be 343 m/s.) ___m/s (b) What is its altitude? ___m

When the temperature of the ocean is 80°F what is likely to develop is Hurricane. That is option D.

Hurricane is a natural disaster that occurs over the tropical large water bodie such as the oceans which is characterized by low-pressure storm.

The factors that can cause the occurrence of hurricane include the following;

Therefore, when the temperature of the ocean is 80°F what is likely to develop is Hurricane.

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Answer: thunderstorm

Explanation:

When the sea surface is at least 80° Fahrenheit (27° Celsius), it can supercharge a thunderstorm. The storm sucks up that heat and water, which make the storm bigger. As it grows, air pressure at the center of the storm continues to drop, which causes the vacuum in the middle to grow stronger.

Answer:

Explanation:

Well they told you the exact formula to use. Work is the force multiplied by  the distance through which its applied.

W = (20N)(35m)

= 700 Joules

13.) Power is the amount of work done over the time through which the work is being done.

P = W/t

= 10J/2s

= 5J/s

Answer:

i believe its 3

Explanation:

An airplane is flying with a force of 800 N. It experiences a force of air resistance of 40 N and a wind force of 60 N, both acting in the opposite direction that the plane is traveling. The net force on the airplane will be 700 N

Force with which airplane is flying ( F1 ) = 800 N

opposite to which two forces are acting

air resistance = 40 N

wind force = 60 N

net force = F1 - 40 - 60 = 800 - 100 = 700 N

The net force on the airplane will be 700 N

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Answer:

Explanation:

The answer is **(c) 9.39 m/s** for the velocity of the bullet-block system after it's hit, and **(g) 209.39 m/s** for the bullet's velocity before it hit the box.

The velocity of the bullet-block system after it's hit can be found using the conservation of energy. The potential energy of the box before it was hit is mgh, where m is the mass of the box, g is the acceleration due to gravity, and h is the height that the box was displaced. After the bullet hits the box, the potential energy of the box is zero, but the kinetic energy of the bullet-block system is mv^2/2, where m is the total mass of the bullet-block system and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mgh = mv^2/2

```

Solving for v, we get:

```

v = sqrt(2mgh)

```

In this case, we have:

* m = 0.05 kg + 1.3 kg = 1.35 kg

* g = 9.8 m/s^2

* h = 4.5 m

So, the velocity of the bullet-block system after it's hit is:

```

v = sqrt(2 * 1.35 kg * 9.8 m/s^2 * 4.5 m) = 9.39 m/s

```

The velocity of the bullet before it hit the box can be found using the conservation of momentum. The momentum of the bullet before it hit the box is mv, where m is the mass of the bullet and v is the velocity of the bullet. After the bullet hits the box, the momentum of the bullet-block system is (m + M)v, where M is the mass of the box and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mv = (m + M)v

```

Solving for v, we get:

```

v = mv/(m + M)

```

In this case, we have:

* m = 0.05 kg

* M = 1.3 kg

* v = 9.39 m/s

So, the velocity of the bullet before it hit the box is:

```

v = 0.05 kg * 9.39 m/s / (0.05 kg + 1.3 kg) = 209.39 m/s

```

The velocity of the bullet-block system after the collision is approximately a) 6.76 m/s, and the bullet's velocity before it hit the box is approximately e) 196.76 m/s.

To solve this problem, we can apply the principle of conservation of momentum and the principle of conservation of mechanical energy.

First, let's calculate the velocity of the bullet-block system after the collision. We can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Let m1 be the mass of the bullet (0.05 kg) and m2 be the mass of the box (1.3 kg). Let v1 be the velocity of the bullet before the collision (which we need to find) and v2 be the velocity of the bullet-block system after the collision.

Using the conservation of momentum:

m1 * v1 = (m1 + m2) * v2

0.05 kg * v1 = (0.05 kg + 1.3 kg) * v2

0.05 kg * v1 = 1.35 kg * v2

Now, let's calculate the velocity of the bullet-block system (v2). Since the system goes up by a height of 4.5 m, we can use the principle of conservation of mechanical energy.

m1 * v1^2 = (m1 + m2) * v2^2 + m2 * g * h

0.05 kg * v1^2 = 1.35 kg * v2^2 + 1.3 kg * 9.8 m/s^2 * 4.5 m

Now, we can substitute the value of v2 from the momentum equation into the energy equation and solve for v1.

By solving these equations, we find that v1 is approximately 196.76 m/s.

Therefore, the bullet's velocity before it hit the box is approximately 196.76 m/s. (e) 196.76 m/s

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The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

The average speed of the ant is calculated as;

\(average \ speed = \frac{total \ distance }{total \ time }\)

The total distance from the graph is calculated as follows;

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

\(average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s\)

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

Notice, you have a right triangle, now calculate the length of  line P.

                                                ↓end

                                                ↓

                                                ↓ 6cm

                                                ↓

  start -------------13 cm------------

Use Pythagoras theorem to solve for P.

\(P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm\)

The average velocity of the ant is calculated as;

\(average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\\)

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

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Answer:16

Explanation:

as F=BIL so by placing value's B is magnetic field 4 and i is current 2A and l is length that js 2 m so answer is

Given: i=10A,B=0.15 T,θ=45

∘

and l=2 m

Force on a current carrying wire of finite length =

F

=l(

i

×

B

)

Hence, magnitude of force, F=Bilsinθ

=0.15×10×2×sin45

∘

=

2

3

N

Answer:

A dump truck.

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

Mathematically, it is given by the formula;

\( F = ma\)

Where;

The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a dump truck, the greater will be its tendency to continuously remain at rest.

Hence, a dump truck would probably need the greatest force to overcome its inertia because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia, as well as the force applied.

Answer:As one of Newton's Laws of Gravitational force The answer would Be a dump truck because it includes one of Newton's Laws of Gravitational force in the abstract Laws of gravity.

Answer:

the answer is electromagnetic force because gravity is a force of attraction

The product of the change in velocity and the change in mass is the change in momentum. Because its value is constant, the mass does not change.

Δp = m Δvwhere p is the change in momentum, m is the mass, and v is the change in the object's velocity.

The ball flies off in the opposite direction from where it came from, as is typical of ball-bat collisions. If we make the bat weigh 25 ounces, if we swing it at the same speed, we get a significant boost in bat momentum, which raises the ball's exit velocity.

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Answer:

Option (b) is correct.

Explanation:

An object travels down a ramp at a constant acceleration.

Friction force and the gravitational force is there.

(a). It is false. As the friction acts in the opposite direction of motion of the object.

(b). It is true. As the object moves down the ramp so the force of friction is less the component of gravitational force along the ramp.

(c). It is false. the acceleration does not depends on the mass.

a = g sin A - u g cos A

where, A is the angle of inclination and u is the coefficient of friction.

(d). It is false. The normal force is given by

N = m g cos A

so it depends on the mass.

Answer:

6767

Explanation:

Answer:

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA

Initial diameter of the wire, d₁ = 4 mm = 0.004 m

Final diameter of the wire, d₂ = 1 mm = 0.001 m

Length of wire, L = 2.00 m

Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

The final area of the copper wire;

The initial drift velocity of the electrons is calculated as;

The final drift velocity of the electrons is calculated as;

The change in the mean drift velocity is calculated as;

The time of motion of electrons for the initial wire diameter is calculated as;

The time of motion of electrons for the final wire diameter is calculated as;

The average acceleration of the electrons is calculated as;

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

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Explanation:

The changing mean drift velocity of the electrons plays out at 3.506 x 10⁻⁷ m/s along with an average acceleration nearing 4.38 x 10⁻¹⁵ m/s².

As the electrons traverse one end of the wire to another, their mean drift velocity undergoes a shift of 3.506 x 10⁻⁷ m/s with an average acceleration of 4.38 x 10⁻¹⁵ m/s² in accordance with the following parameters:

- The current flowing through the wire is at 4.00 mA.

- The original diameter of the wire, d₁, measures at 4 mm or 0.004 m.

- Conversely, the final diameter, d₂, displays a measurement of 1 mm or 0.001 m.

- The length of the entire wire is consistent, measuring at 2 meters.

- Notably, the density of electrons present within copper reaches an estimated value of 8.5 x 10²⁸ /m³.

Calculations regarding both initial and final area coverage provided by copper must be explored along with numerical data involving the two varying drift velocities for accurate results.

Thus, we arrive at the change rate of the mean drift velocity between points in the wire as well as the plenitude of electron acceleration achieved after contemplation into the corresponding motion periods.

The conclusion reflects that our measurements find the changing mean drift velocity of the electrons plays out at 3.506 x 10⁻⁷ m/s along with an average acceleration nearing 4.38 x 10⁻¹⁵ m/s².

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Answer:

Approximately 2*10^13 electrons must be transferred

Explanation:

Below is the given information:

First object carries charge = -8.5 µC

Number of electrons in 1st = 8.5 x 10^-6/(1.6 x 10^-19) = 5.3125 x 10^13

Second object carries a charge = -2.0 µC

The number of electrons in 2nd = 2*10^-6/(1.6*10^-19) = 1.25 x 10^13

so, approximately 2 x 10^13 electrons must be transferred

Answer:

Explanation:

The electric flux through a surface is given by the dot product of the electric field and the area vector of the surface:

Φ = E · A

where Φ is the electric flux, E is the electric field, and A is the area vector of the surface.

(a) If the surface lies in the yz plane, its area vector is in the x direction. Therefore, the area vector can be written as A = Ax i, where Ax is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Ax i) = aAx

(b) If the surface lies in the xz plane, its area vector is in the y direction. Therefore, the area vector can be written as A = Ay j, where Ay is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Ay j) = bAy

(c) If the surface lies in the xy plane, its area vector is in the z direction. Therefore, the area vector can be written as A = Az k, where Az is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Az k) = 0

since the dot product of perpendicular vectors is zero.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer: They will repel each other.

Explanation:

Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.

Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.

As, we know that the same charges repeal each other both of the balloons with be apart from each other.

This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.

Answer:

Sample Response: Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.

Explanation:

Did it on Egde 2020

The electric potential energy stored in the capacitor is 4.70 x 10^-8 Joules.

The electric potential energy stored in a capacitor is given by the formula:

U = (1/2) * C * V^2

where U is the potential energy in Joules, C is the capacitance in Farads, and V is the voltage across the capacitor in Volts.

In this case, we are given that the capacitor stores 9.40 x 10^-10 C of charge at 50.0 V. However, we are not given the capacitance value. Therefore, we cannot calculate the potential energy directly using the above formula.

To find the capacitance value, we can use the formula:

C = Q / V

where Q is the charge stored in the capacitor and V is the voltage across the capacitor.

Substituting the given values, we get:

C = 9.40 x 10^-10 / 50.0

= 1.88 x 10^-11 F

Now we can use the formula for electric potential energy to find the energy stored in the capacitor:

U = (1/2) * 1.88 x 10^-11 * (50.0)^2

= 4.70 x 10^-8 J

Therefore, the electric potential energy stored in the capacitor is 4.70 x 10^-8 Joules.

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Answer: the answer is 19cm

Explanation:

The velocity of the wave at point S, given the speed and the distance change, is 6 cm/s.

The wavelength of the wave at point R is 3. 0 cm, and the frequency of the wave is:

= 12 cm / s / 3. 0 cm

= 4 Hz

The wavelength of the wave at point S is 1. 5 cm, and the frequency of the wave is still 4 Hz .

The velocity of the wave at point S is therefore:

= 4 Hz x 1. 5 cm

= 6 cm / s

In conclusion, the velocity of the wave at point S is 6 cm / s.

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The combination of two or more wave trains travelling down the same or adjacent paths results in a wave train effect.

The resultant wave will have sum of the individual wave amplitudes at each place that is influenced by multiple waves.

The undulating fluctuation in forward pressure on the molecules along a sound wave's journey is known as a pressure wave train.

The relative shift in a wave's position that occurs when it passes through a medium is referred to as wave displacement.

Phase of the pressure wave is π/2 times the displacement wave.

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Answer:

Explanation:

The picture shown in the figure represents the Milky Way Galaxy. The Galaxy in which the entire solar system is present.

The million and trillion of stars in the universe form Galaxy. The galaxy in which the entire solar system is present is called Milky Way Galaxy. The Milky Way Galaxy is spiral in shape. This Galaxy has four major arms. The major arms have both old and young stars and the minor arms have the gas and star formation activity. This galaxy also has a black hole at its center. Galileo Galilei was the first to see the Galaxy.

The Milky Way Galaxy is made up of a dense cloud of gas that stretches across the sky as seen from the Earth. The age of the Milky Way Galaxy is 13.61 billion years ago and the Andromeda Galaxy is the nearest galaxy to the Milky Way Galaxy.

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Answer:

500 Hz

Explanation:

Formula for finding wave frequency is,

f = c/λ

f = frequency

c = speed (m/s)

λ = wave length (m)

f = c/λ

f = 280/0.56

f = 500

∴ wave frequency is 500 Hz

Answer:

Explanation:

a = gsinθ = 9.8sin30 = 4.9 m/s²

When you change the light from green to red, the number of waves remains the same.

Light waves behave similarly throughout the electromagnetic spectrum. Depending on the nature of the item and the light's wavelength, a light wave can be transmitted, reflected, absorbed, refracted, polarized, diffracted, or dispersed when it strikes a surface.

The number of waves doesn't vary when the light changes from green to red. The distance between each wave's consecutive peaks, or wavelength, determines the color of light. Red and green light both have waves with a set number of peaks and troughs per unit of time, despite green light having a shorter wavelength. As a result, the number of waves is unaffected by changing the colour of the light.

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A.  The motor's speed is approximately 1086 r/min if the field resistance is raised to 50 Ω.

B. The speed of this motor as a function of the field resistance RF is approximately 1086 r/min

A. According to the magnetization curve shown in Figure 8-9, the motor's speed can be calculated by using the following equation:

EA = kϕN, where EA is the back EMF, k is a constant, ϕ is the magnetic flux, and N is the motor speed.

Since the amount of armature current remains constant, the back EMF is also constant.

Therefore, the magnetic flux must also be constant. The magnetic flux is proportional to the field current IF, which can be calculated using Ohm's law:

IF = (250 V - EA)/(RF + R₁)

At the initial field resistance of 41.67 Ω, the field current is IF = (250 V - EA)/(41.67 Ω + 0.03 Ω) = (250 V - EA)/41.70 Ω.

If the field resistance is raised to 50 Ω, then the new field current is IF = (250 V - EA)/(50 Ω + 0.03 Ω) = (250 V - EA)/50.03 Ω.

Since the magnetic flux is constant, we can set the two expressions for IF equal to each other and solve for N:

kϕN/IF1 = kϕN/IF2

N = (IF2/IF1)N1 = (250 V - EA)/(50.03 Ω + 0.03 Ω) * 1103 r/min ≈ 1086 r/min

Therefore, the motor's speed is approximately 1086 r/min if the field resistance is raised to 50 Ω.

B. The speed of the motor as a function of the field resistance RF can be plotted using the same equation used in part A:

N = (250 V - EA)/(RF + R₁ + Radj) * 1103 r/min

where Radj is the resistance of any additional resistance in the circuit. Since the load current is constant, the current through the motor is also constant, so EA is also constant.

Therefore, the speed is inversely proportional to the total resistance in the circuit, which includes the field resistance RF, armature resistance R₁, and any additional resistance Radj.

A plot of the speed as a function of the field resistance is shown in Figure 8-10. As the field resistance increases, the speed of the motor decreases due to the increased total resistance in the circuit. This relationship is linear for this type of constant-current load.

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Answer:

  (possibly) Box D

Explanation:

The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.