--[50 POINTS]--
1)A block of mass 25 kg is placed on flat ground. The coefficient of static friction and kinetic friction are 0.73 and 0.16
a.If a person pushes the block and the block is moving, what will be the acceleration of the block?


2) A block has a mass of 79 kg. The coefficient of static and kinetic friction between the sled and the ground is 0.87 and 0.37. Person A tries to pull the block with 210N, but fails.
a) Person B successfully pulls the sled with 909N. What is the acceleration of the sled?

The magnetic field at a point due to wires with distance r = C cm and carrying the current I is 2 mA is 0.29 T.

The magnetic field is produced over the magnet or current-carrying conductor. The Biot-Sarvat law is defined as the magnetic field produced due to the current-carrying conductor. The magnetic field is the vector quantity and the unit of the magnetic field is Tesla. It is the vector quantity.

Magnetic field, B = μ₀/2π (I/R), where  μ₀ is the permeability of the magnetic field, I is current, and R is the distance between two points.

B₁= μ₀/2π (I/R)

 = (1.26×10⁻⁶)×2×10⁻³/(2π×1.35m)

 = 0.29T

B₂ = 0.29T

The net magnetic field is B=B₁+B₂=0.58T.

Thus, the magnetic field is 0.29T. The left wire carries the current outward direction and the right wire carried the current inward direction and hence the direction magnetic field given by the right-hand thumb rule is inside.

B) By using the equation, F=q(v×B), where q is the charge, v is the velocity and B is the magnetic field. Path b is followed by a neutron as it is not deflected by a Magnetic field. The path a is the upward path followed by a negatively charged particle(electron) and Path C is the downward path followed by a positively charged particle(proton).

C) The circular conductor undergoes thermal expansion while it is a uniform magnetic field the direction of the current is a clockwise direction given by the Right-hand thumb rule.

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Based on the principle of moments and the data provided;

The principle of moments states that the for a body in equilibrium, the sum of the clockwise moments is equal to the sum of the anticlockwise moments about a point.

At the point where the uniform rod is about to tilt, it is at equilibrium.

Anticlockwise moments = 30 × (1.3 - 0.5) = 24

Clockwise moments = m × (2.5 - 1.3)

where m is the mass of the end

Clockwise moments = m × 1.2

1.2 × m = 24

m = 20 Kg

Therefore, the mass of the rod is 20 kg

Assuming the boy moves to B and a block of mass X is placed at:

Anticlockwise moments = 30 × 1.3 = 39

Clockwise moments = 20 × 1.2 + X × 3.7

Clockwise moments = 24 + 3.7X

39 = 24 + 3.7X

3.7X = 15

X = 4.054 kg

Therefore, the smallest value of X of the block that will enable the boy to stand on the beam at B without the beam tilting is 4.054 kg

Assuming that the block is a particle allowed the assumption that the block touches the rod at only a point.

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Answer: The ball reaches a height of 68 feet after 1.7 seconds.

Explanation: To solve this problem, we can use the following formula:

h(t) = -16t^2 + vt + h

where h(t) is the height of the ball at time t, v is the initial velocity (in this case, 80 ft./sec.), and h is the initial height (in this case, 4 feet).

We want to find out when the height of the ball is 68 feet, so we can set h(t) = 68 and solve for t:

68 = -16t^2 + 80t + 4

Rearranging and simplifying:

16t^2 - 80t - 64 = 0

Dividing both sides by 16:

t^2 - 5t - 4 = 0

Factoring:

(t - 4)(t + 1) = 0

So t = 4 or t = -1. We can discard the negative value, since time cannot be negative. Therefore, the ball reaches a height of 68 feet after t = 4 seconds - the time at which it reaches maximum height.

However, we also know that the initial height of the ball is 4 feet, so we need to subtract this from the maximum height to get the actual height above the ground:

max height = h(4) = -16(4)^2 + 80(4) + 4 = 132 feet

So the ball first reaches a height of 68 feet on its way down, after reaching a maximum height of 132 feet. We can use the formula again to find when the ball reaches 68 feet on its way down:

68 = -16t^2 + 80t + 132

Rearranging and simplifying:

16t^2 - 80t + 64 = 0

Dividing both sides by 16:

t^2 - 5t + 4 = 0

Factoring:

(t - 1)(t - 4) = 0

So t = 1 or t = 4. We discard the value t = 4, since that is when the ball reaches maximum height. Therefore, the ball reaches a height of 68 feet again, on its way down, after t = 1 second.

Therefore, the ball reaches a height of 68 feet after 1.7 seconds (1 second on the way down + 0.7 seconds to reach maximum height on the way up).

Hope this helps, and have a great day!

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

Substitute these values into equation 2

e = 10/395

e = 0.025 m

Answer:

Answer choices

Accuracy in the measurements

Significant digits in the measurements

Precision in the measurements

He goes 400 m straight then goes 500m to the left side then goes backwards 400m so walking north is cancelled out he covers only 500 m to the west.

If u mean how much distance he travelled from starting point then this is the answer

Answer: He goes 400 m straight then goes 500m to the left side then goes backwards 400m so walking north is cancelled out he covers only 500 m to the west.

If you mean how much distance he travelled from starting point then this is the answer

Your acceleration varies as you travel to the store. Your total speed can change over time; sometimes it can go up, sometimes down. Because doing so would mean you would keep speeding up and eventually pass the store, constant acceleration doesn't happen. In physics, even a decrease in speed is a form of acceleration. When you arrive at it, we can determine how much your average speed was and examine the impact that varying acceleration would have had on it.

According to estimates of the power needed to ride, cyclists deliver about 150 W when riding level terrain, but this increases to about 250 W when cycling up hills. On a flat surface, the average acceleration is 0.231 m/s2, and the acceleration phase has an average duration of 26 seconds. On average, 120 W of power is output during this phase.

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The ions are;

S^2- Gained 2 electrons

Li^+ - Lost one electron

Al^3+ - Lost 3 electrons

Cl^- gained one electron

Sr^2+ Lost two electrons

P^3- gained 3 electrons

As an atom or molecule acquires or loses one or more electrons, ions are created. Protons and neutrons make up the positively charged nucleus of an atom, which is encircled by negatively charged electrons. An atom typically has a neutral charge because the amount of electrons and protons in it equals one.

The balance between the negatively charged electrons and positively charged protons is upset when an atom receives or loses an electron, creating an ion with a positive or negative charge.

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Answer:

ρ (density) = M / V          mass / volume

V = M / ρ = 19 kg / 800 kg/m^3 = .024 m^3

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.  

1. The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 8% or 0.08.

2. The final kinetic energy of the neutron KE₂ (neutron), is 6.42 x 10⁻¹³ J.

Kinetic energy is the energy possessed by a body by virtue of its motion.

The fraction of the neutron's kinetic energy that is transferred to the atomic nucleus is calculated as follows:

The final velocity of the atom is found using the principle of conservation of linear momentum.

initial momentum of the neutron = final momentum of the atom

m₁u₁  = m₂u₂

where;

m₁ = mass of the neutron

u₁ = initial velocity of the neutron

m₂ = mass of the atomic nucleus

u₂ = final velocity of the atomic nucleus

m₂ = 112.4 m₁

u₂  = m₁u₁ / m₂

u₂  = m₁u₁ / (12.4 m₁)

u₂  = 0.08 u₁

The initial kinetic energy of the neutron is determined as follows:

KE₁ = ¹/₂m₁u₁²

The final kinetic energy of the atomic nucleus is determined as follows:

KE₂ =  ¹/₂m₂u₂²

KE₂ =  ¹/₂(12.4 m₁)(0.08u₁)²

KE₂  = 0.08 (¹/₂m₁u₁²)

KE₂ = 0.08 (KE₁)

The fraction of the neutron's kinetic energy transferred to the atomic nucleus is determined as follows:

Fraction = 0.08 (KE₂) / KE₁

Fraction= 0.08

Fraction = 8 %

The final kinetic energy of the neutron is calculated as follows;

KE₂ (neutron) = (1 - 0.08) x (6.97 x 10⁻¹³ J)

KE₂ (neutron) = 6.42 x 10⁻¹³ J

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So, the force of gravity that the asteroid and the planet have on each other approximately

\( \boxed{\sf{5.43 \times 10^{10} \: N}} \)

Hi ! Now, I will help to discuss about the gravitational force between two objects. We already know that gravitational force occurs when two or more objects interact with each other at a certain distance and generally orbit each other to their center of mass. For the gravitational force between two objects, it can be calculated using the following formula :

\( \boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}} \)

With the following condition :

We know that :

What was asked :

Step by step :

\( \sf{F = G \times \frac{m_1 \times m_2}{r^2}} \)

\( \sf{F = 6.67 \times 10^{-11} \times \frac{8.4 \cdot 10^8 \times 6.2 \cdot 10^{23}}{(8 \times 10^5)^2}} \)

\( \sf{F = \frac{347.374 \times 10^{-11 + 8 + 23}}{64 \times 10^10}} \)

\( \sf{F \approx 5.43 \times 10^{20 - 10}} \)

\( \boxed{\sf{F \approx 5.43 \times 10^{10} \: N}} \)

So, the force of gravity that the asteroid and the planet have on each other approximately

\( \boxed{\sf{5.43 \times 10^{10} \: N}} \)

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Answer:

About 6 feet or higher is recommended.  

Explanation

Answer:

What are igneous rocks?

Igneous rocks (from the Latin word for fire) form when hot, molten rock crystallizes and solidifies. The melt originates deep within the Earth near active plate boundaries or hot spots, then rises toward the surface. Igneous rocks are divided into two groups, intrusive or extrusive, depending upon where the molten rock solidifies

Answer:   Mechanical energy is the energy that is possessed by an object due to its motion or due to its position.

(The energy acquired by the objects upon which work is done)

Answer:

The International Lawn Tennis Federation, now known simply as the International Tennis Federation, the sport's governing body, was founded in 1913, composed of 13 national tennis associations. The Modern Olympics , growing out of the ancient tradition, resurfaced under the direction of the International Olympic Committee in Athens in 1896.

Explanation:

4.44×\(10^{9}\) kg/s is the rate at which the sun mass is decreasing.

The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.

To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.

Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.

ΔE = 4×10^26 W

c = 3×10^8 m/s (speed of light)

Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².

Substituting the values, we have:

Δm = (4×10^26 W) / (3×10^8 m/s)²

Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.

This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.

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Answer:

Explanation:

given:

distance = 270 km

speed = 85 km/h

find:

how long does it take to get into his audition in hours?

solution:

velocity = distance / time

85 km/h =  270 km  

                      t

85 (t) = 270

t = 270 / 85

t = 3.176 hours

The vector field that expresses the density of induced or permanent electric dipole moments in a dielectric medium is known as polarization density (also known as electric polarization or just polarization).

Thus, A dielectric is considered to be polarized when its molecules acquire an electric dipole moment when exposed to an external electric field.

Electric polarization of the dielectric is the term used to describe the electric dipole moment induced per unit volume of the dielectric material.

The forces that emerge from these interactions can be calculated using the polarization density, which also defines how a material reacts to an applied electric field and how it modifies the electric field. It is comparable to magnetization, which measures a material's equivalent response.

Thus, The vector field that expresses the density of induced or permanent electric dipole moments in a dielectric medium is known as polarization density (also known as electric polarization or just polarization).

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The correct statement is Molecular collisions transfer thermal energy between the system and its surroundings. Thus, option D is correct.

The energy moves between a system of reacting substances and the surroundings by the collision of molecules. The transfer of heat or thermal energy between the system and its surroundings by the process of Conduction. Conduction is the process of transmitting the heat to the neighboring atoms or collisions by the process of collisions.

The conduction takes place more steadily in solids and liquids where the molecules are closer together. When the molecules are collided with the nearby molecules, the potential energy is converted into kinetic energy and hence the heat energy is transferred between the system and its surroundings.

Hence, Molecular collisions transfer thermal energy between the system and its surroundings. Thus, the correct option is D.

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Answer:

9.18/13=0.7062 second

Explanation:

9.18meters devide it by the speed 13m/s

≈0.7062

Answer:

The Great Pyramid of Giza (also known as the Pyramid of Khufu or the Pyramid of Cheops) is the oldest and largest of the pyramids in the Giza pyramid complex bordering present-day Giza in Greater Cairo, Egypt.It is the oldest of the Seven Wonders of the Ancient World, and the only one to remain largely intact.

Explanation:

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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Answer:

Its nymber 2

Explanation:

The net force on q₂ will be  1.07 x 10⁻² N, pointing to the left.

To find the net force on particle q₂, we need to calculate the force due to q₁  and q₃ individually and then add them up vectorially. We can use Coulomb's law to calculate the force between two point charges:

F = k × (q₁ × q₂) / r²

where F is the magnitude of the force, k is Coulomb's constant (k = 8.99 x 10⁹ Nm²/C²), q1 and q2 are the charges of the two particles, and r is the distance between them.

The force due to q₁ on q₂ can be calculated as:

F₁ = k × (q₁ × q₂) / r₁²

where r1 is the distance between q₁ and q₂ (r₁ = 0.180 m).

Similarly, the force due to q₃ on q₂ can be calculated as:

F₂ = k × (q₃ × q₂) / r₃²

where r₃ is the distance between q₂ and q₃ (r₃= 0.230 m).

The direction of each force can be determined by the direction of the electric field due to each charge. Since q₁ and q₃ have opposite signs, their electric fields point in opposite directions. Therefore, the force due to q₁ points to the left and the force due to q₃ points to the right.

To find the net force, we need to add up the forces vectorially. Since the forces due to q₁ and q₃ are in opposite directions, we can subtract the magnitude of the force due to q₃ from the magnitude of the force due to q₁ to get the net force on q₂:

Fnet = F₁ - F₃

Substituting the values we get:

Fnet = k × (q₁ × q₂) / r₁² - k × (q₃ × q₂) / r₃²

Plugging in the values we get:

Fnet = (8.99 x 10⁹ Nm²/C²) × [(9.33 x 10⁻⁶ C) × (4.22 x 10⁻⁶ C) / (0.180 m)² - (-8.42 x 10⁻⁶ C) × (4.22 x 10⁻⁶ C) / (0.230 m)²]

Fnet = 1.07 x 10⁻² N

Therefore, the net force on q₂ is 1.07 x 10⁻² N, pointing to the left.

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Explanation:

13x - 26 + 39 = 0

13x=26-39

13x=-13

x= -1

\(13x-26+39=0\\\\\implies 13x +13 = 0\\\\\implies 13x = -13\\\\\implies x = -\dfrac{13}{13}\\\\\implies x = -1\)

To determine the magnetic flux density at point P, we can use the Biot-Savart Law. The law states that the magnetic field at a point due to a current-carrying wire is proportional to the current and the distance from the wire.

Since we have two parallel wires carrying equal and opposite currents, the magnetic fields they produce will cancel each other out at a point equidistant from them. Thus, the magnetic field at point P is zero.

Alternatively, we can use the right-hand rule to determine the direction of the magnetic field produced by each wire. If we point the thumb of our right hand in the direction of the current in wire 1, and the fingers in the direction of wire 2, the palm will face the direction of the magnetic field at point P. Since the currents are in opposite directions, the magnetic fields produced by the wires will be in opposite directions, and their vector sum at point P will be zero. Therefore, the magnetic flux density at point P is zero.

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