The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizontal circle with a radius of 2.5 m, at a constant speed of 8.6 m/s. The floor is lowered and the student remains against the wall without falling to the floor.
The magnitude of the centripetal force acting on the student at point A is approximately
Question 1 options:

A.) 1923 N
B.) 47 N
C.) 1398 N
D.) 223 N

The tourist is standing one-third of the way along the bridge (option c).

Let's denote the distance from the near end of the bridge to the point where the tourist is standing as x and the total length of the bridge as L.

According to the equilibrium condition, the sum of the forces exerted on the bridge is zero.

So, the vertical forces exerted by the two concrete supports should be equal to the weight of the tourist:

F_near + F_far = 705 N

Given that the magnitude of the vertical force exerted by the near end support is 470 N, we can calculate the force exerted by the far end support:

F_far = 705 N - 470 N = 235 N

Now, we can use the moment equilibrium condition, considering moments around the near end support:

Moment = Force × Distance

For the tourist:

Moment_tourist = 705 N × x

For the far end support:

Moment_far = 235 N × L

For equilibrium, the sum of the moments should be zero:

Moment_tourist - Moment_far = 0

Substituting the moments:

705 N × x - 235 N × L = 0

Now, we can solve for x/L, which represents the fraction of the way along the bridge where the tourist is standing:

x/L = (235 N × L) / 705 N

x/L = 235/705 = 47/141 ≈ 1/3

So, the tourist is standing one-third of the way along the bridge (option c).

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The weight of 15N should be hung at a distance of 4/3 units from the fulcrum, on the opposite side of the bar to a weight of 20N.

The fulcrum is attained equilibrium when the principle of moments is equal to zero. The principle of momentum is defined as the body is said to be balanced when the clockwise movement about the point is equal to the anticlockwise movement about the same point.

When a weight of 15N is hung on the bar, it produces an anticlockwise movement with the same magnitude, opposes the clockwise moment produced by the weight of 20N on the fulcrum.

To find the distance (d) at which the 15N weight should be hung, the distance from the 20N weight to the fulcrum is 1 unit.

20×1 = 15×d

d = 20/15

  =4/3 m

Thus, the weight of 15N should be hung by the distance of 4/3 units from the fulcrum, on the opposite side of the bar having the weight of 20N, to make the fulcrum to be at equilibrium.

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The speed  in meters per second if the string breaks suddenly is 9.6 m/s.

The angular speed is the speed of an object that is moving along a circular path. In this case, we have a puck of mass m = 0.095 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.69 m. The puck makes one revolution every t = 0.45 s.

Let us recall that in order to find the angular speed as is required by the question "how fast  in meters per second, does the puck move away?" we have to get the angle turned as well as the time taken.

We are told here that the angle turned is one revolution  or 2π radians and the time from the question is 0.45 s. Thus the speed is gotten from;

ω = θ/t

ω = 2π radians/0.45 s

ω = 13.96 rad/s

But v = rω

v = 13.96 rad/s * 0.69 m

v = 9.6 m/s

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Answer:

2 Pascal (Pa)

Explanation:

Pressure is defined as the force acting per unit area. Mathematically, it is expressed as:

Pressure (P) = Force (F) / Area (A)

Given:

Force exerted by the air inside the balloon (F) = 1 N

Area of the balloon (A) = 0.5 m^2

Plugging in the given values into the formula for pressure, we get:

P = F / A

P = 1 N / 0.5 m^2

Using basic arithmetic, we can calculate the pressure inside the balloon:

P = 2 N/m^2

So, the pressure inside the balloon is 2 N/m^2, which is also commonly referred to as 2 Pascal (Pa) since 1 Pascal is equal to 1 N/m^2.

The thermosphere has the highest temperature of any layer in Earth’s atmosphere.

The term troposphere is the region that is found  12 to 50 kilometers from Earth’s surface. This region is found to be the region where you can find a lot of gases.

Both the Troposphere and the stratosphere  get colder as altitude increases. However, the ozone in the stratosphere  protects people from ultraviolet (UV) radiation.

The thermosphere has the highest temperature of any layer in Earth’s atmosphere.

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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: I am so sorry i hadn't learned this Yet~!

Explanation:

Creativity has been studied in psychology. First, it ranges from individuals with little to no original thought to very creative and innovative thinkers. It does not just take the form of art. Consider the industrial titans Edison, Ford, and Harvey Firestone, all of whom were also extremely creative people. They are equally as imaginative as Beethoven, Monet, and Picasso. One can be creative in many different ways or just one. There might be a biological or genetic foundation for creativity. Take a look at the Bach family, Paloma Picasso, Eugen, and Manfred Bleuler for evidence that it may be inherited to some extent (father and son psychiatrists who greatly advanced the study of schizophrenia). After saying that, I have to add that creativity also requires HARD WORK, PRACTICE, etc. The traditional quip that just 10% of creativity comes from heredity or hard effort is actually accurate. I apologize if I misunderstood the humor. Before discovering the right mix of materials, Edison would test 1,000 different combinations. You can find piles of canvasses, shattered pots, and other items that failed to live up to the artist's expectations in their studio. The ability to persist is crucial to creativity. Divergent thought is yet another feature. Most individuals only see one solution when they examine a situation. When someone is creative, they might imagine dozens or even hundreds of solutions to the same issue. People that are creative don't hesitate to attempt and reject ideas.  The creative process overvalues intelligence. The link between intellect and creativity disintegrates over a relatively low degree. And there are some fairly unimpressive individuals who are incredibly creative but not particularly gifted at managing their finances or producing academic work. To some extent, creativity may be taught. We must begin early in the learning and growth process in order to be as effective as possible.

Science knows quite a bit about creativity—but there remains much to learn.

- Eddie

Answer:

The centripetal acceleration (ac)=314m/s²

Explanation:

look at the attachment ☝️

Answer:

pictures please

Explanation:

I need a picture so I can tell you

Answer:

55.897905

Explanation:

1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth

9.80665 times 5.7=55.897905

Brainliest?

Answer:206.3

Explanation:

Answer:

New technology allows scientists to explore different aspects of their theories. It also allows them to explore their theories in greater depths.

Motion can be measured in several ways, depending on the context and the specific parameters that need to be quantified.

What is Motion?

Motion is a fundamental concept in physics and is described by the laws of motion formulated by Sir Isaac Newton. These laws state that an object at rest tends to remain at rest and an object in motion tends to remain in motion with the same velocity . The study of motion is important in many fields, including physics, engineering, and astronomy.

Motion can be measured in several ways, depending on the context and the specific parameters that need to be quantified. Some common methods of measuring motion include:

Displacement: Displacement is the change in position of an object over a certain period of time. It can be measured using a ruler, tape measure, or other distance-measuring tools.

Speed: Speed is the rate at which an object moves, and it can be measured using a stopwatch or other timing device to determine the time it takes for an object to travel a certain distance.

Velocity: Velocity is similar to speed, but it takes into account both the speed and direction of motion. It can be measured using a velocity sensor or other motion-tracking tools.

Acceleration: Acceleration is the rate at which an object's speed or velocity changes over time, and it can be measured using an accelerometer or other motion-tracking tools.

Gyroscopic sensors: Gyroscopic sensors can be used to measure the rotational motion of an object, such as the rotation of a wheel or the movement of a drone.

Video analysis: Video analysis software can be used to track the movement of an object or a person, using visual markers or other reference points to determine motion.

Inertial measurement units (IMUs): IMUs are electronic sensors that can measure motion, acceleration, and rotation in three-dimensional space, and are commonly used in robotics, virtual reality, and other applications.

These are just a few examples of methods used to measure motion, and there are many other tools and techniques that can be used depending on the specific context and requirements.

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Answer:

Vy = g T       where T is the time taken to fall

1 = 1/2 g T^2      time to fall 1 m

T = (2 / 9.8)^1/2 = .45 sec

Vy = 9.8 * .45 = 4.43 m/s

V = (Vx^2 + Vy^2)^1/2 = (16 + 19.6)^1/2 = 5.97 m/s

.

The weight of the aluminum foil is 26.20 N

To find the weight of the aluminum foil when it is unrolled, we need to find its volume.

The volume V = lwt where

So, V = lwt

= 22 m × 0.30 m × 0.15 × 10⁻³ m

= 0.99 × 10⁻³ m³.

So, its weight W = ρV where

So, W = ρV

W = 26460 N/m³ × 0.99 × 10⁻³ m³

W = 26195.4 × 10⁻³ N

W = 26.1954 N

W ≅ 26.20 N

So, the weight of the aluminum foil is 26.20 N

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Answer:

(i) To find the velocity of the crate just after losing contact with the roof, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and s is the displacement (which is 3 m).

Substituting the values, we get:

v^2 = 0 + 2 × 4.9 × 3

v^2 = 29.4

v = √29.4

v ≈ 5.42 m/s

Therefore, the velocity of the crate just after losing contact with the roof is approximately 5.42 m/s.

(ii) To find the velocity just before the crate hits the ground, we can use the conservation of energy principle. The initial potential energy of the crate at the top of the roof is converted into kinetic energy just before it hits the ground.

Initial potential energy = mgh

where m is the mass of the crate, g is the acceleration due to gravity, and h is the height of the roof.

Final kinetic energy = (1/2)mv^2

where v is the velocity just before the crate hits the ground.

Since there is no loss of energy, we can equate these two values:

mgh = (1/2)mv^2

Canceling out m on both sides and substituting the values, we get:

(1/2) × 1.17 × 9.81 × 3 = (1/2) × 1.17 × v^2

v^2 = 34.5

v = √34.5

v ≈ 5.87 m/s

Therefore, the velocity just before the crate hits the ground is approximately 5.87 m/s, and its direction is along the horizontal.

(iii) To find the time taken by the crate to hit the ground after leaving the roof, we can use the equation:

s = ut + (1/2)at^2

where s is the displacement (which is the height of the roof, 3 m), u is the initial velocity (which is zero), a is the acceleration, and t is the time taken.

Substituting the values, we get:

3 = 0 + (1/2) × 9.81 × t^2

t^2 = 0.612

t ≈ √0.612

t ≈ 0.78 s

Therefore, the time taken by the crate to hit the ground after leaving the roof is approximately 0.78 s.

(iv) Finally, to find the horizontal distance between the point directly below the roof and the landing point, we can use the equation:

s = ut + (1/2)at^2

where s is the horizontal distance, u is the initial velocity (which is zero), a is the acceleration (which is zero, since there is no force acting along the horizontal direction), and t is the time taken (which is 0.78 s).

Substituting the values, we get:

s = 0 + (1/2) × 0 × (0.78)^2

s = 0

Therefore, the horizontal distance between the point directly below the roof and the landing point is zero, which means the crate lands directly below the roof.

Explanation:

This is quite simple, we can solve this problem using kinematic equations of motion.

(i) The velocity of the crate just after losing contact with the roof:

We can use the kinematic equation, v^2 = u^2 + 2as, where u = 0 m/s (initial velocity), a = 4.9 m/s^2 (acceleration), and s = 3 m (distance).

v^2 = 0^2 + 2(4.9)(3) = 29.4

v = √29.4 = 5.42 m/s (velocity just after losing contact with the roof)

(ii) The velocity (magnitude and direction) just before it hits the ground:

We can use the kinematic equations of motion for motion in a straight line with constant acceleration.

First, let's find the time taken by the crate to reach the ground after leaving the roof. We can use the kinematic equation, s = ut + 1/2 at^2, where s = 3 m (vertical distance), u = 5.42 m/s (initial velocity), a = 9.8 m/s^2 (acceleration due to gravity), and t is the time taken to reach the ground.

3 = 5.42t - 1/2 (9.8)t^2

Simplifying this quadratic equation, we get t = 0.88 s (time taken to reach the ground after leaving the roof).

Now, let's find the velocity just before it hits the ground. We can use the kinematic equation, v = u + at, where u = 5.42 m/s (initial velocity), a = 9.8 m/s^2 (acceleration due to gravity), and t = 0.88 s (time taken to reach the ground after leaving the roof).

v = 5.42 + (9.8)(0.88) = 14.74 m/s (magnitude of the velocity just before hitting the ground)

The direction of the velocity just before hitting the ground is downwards, at an angle of 30 degrees below the horizontal (due to the slope of the roof).

(iii) The time the crate takes to hit the ground after leaving the roof:

We have already calculated this in part (ii), which is t = 0.88 s.

(iv) The horizontal distance between the point directly below the roof and the landing point:

We can use the formula, distance = velocity × time.

The horizontal velocity of the crate is constant and equal to 5.42 m/s, which is the same as the velocity just after losing contact with the roof (since there is no friction).

Therefore, the horizontal distance traveled by the crate before hitting the ground is:

distance = 5.42 m/s × 0.88 s = 4.77 m.

Answer:

velocity= 68.58m/s

Explanation:

formula: v = initial velocity + (acceleration)(time)

plug in your values: v= 18m/s + (1.8 m/s2)(28.1s)

solve the parenthesis first: v = 18 m/s + 58.58m/s

add: v = 68.58 m/s

(i) The direction of the current will be in opposite direction to the magnetic field.

(ii) The magnitude of the emf induced in the circuit is 0.359 V.

(iii) The induced current if the circular coil is 0.048 A.

b(i) The magnetic flux linkage through the coil is 1.41 x 10⁻⁴ Tm².

b(ii) The radius of the coil is r√3 m.

The direction of the current will be in opposite direction to the magnetic field.

Initial area of the coil = (πd²)/4

Initial area of the coil = (π x 0.225²)/4 = 0.0398 m²

Final area of the coil = (π x 0.072²)/4 = 0.001296 m²

\(emf = \frac{NB(A_1 - A_2)}{t} \\\\emf = \frac{14 \times 1.2(0.0398 - 0.001296)}{1.8} \\\\emf = 0.359 \ V\)

I = emf/R

I = 0.359/7.5

I = 0.048 A

Ф = LI

Ф = 0.015 x 0.0094

Ф =  1.41 x 10⁻⁴ Tm²

\(L = \frac{\mu_o N^2\pi r^2}{l} \\\\r^2 = \frac{Ll}{\mu_o N^2\pi } \\\\r = \sqrt{\frac{Ll}{\mu_o N^2\pi } } \\\\r = \sqrt{\frac{0.015(l)}{4\pi\times 10^{-7} \times (420)^2 \pi } } \\\\r = 3 \sqrt{l} \ m\)

where;

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Answer:

The water cycle shows the continuous movement of water within the Earth and atmosphere. ... Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow. Water in different phases moves through the atmosphere (transportation).

Explanation:

It's the water cycle.

Hi there!

We can use work and energy to solve this problem.

We know that:

Ei = Ef

Ei = Potential energy = mgh

Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²

The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:

I (hollow cylinder) = mr²

I (disk) = 1/2mr²

Calculate the moment of inertias of each.

Since the mass on the base is one-fourth of its side:

x = mass of side

x + x/4 = 15

4x + x = 60

5x = 60

x = 12 kg

end mass = 3 kg

Solve for each moment of inertia:

Side: (12)(0.4²) = 1.92 Kgm²

Bottom: 1/2(3)(0.4²) = 0.24 Kgm²

Side + bottom = 2.16 Kgm²

We can now solve:

mgh = 1/2mv² + 1/2(2.16)v²/r²

(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²

4851 = 14.25v²

v = 18.45 m/s

Answer:

The second chart seems to be correct

Explanation:

Part(1)

The voltage across the dependent source is Vx = 1 Volt.

Part(2)

The power supplied by the 4V source is 12 watts.

Kirchhoff's Voltage Law (KVL): The sum of all voltage drops around a closed loop in a circuit must be equal to the sum of all voltage rises around that loop. In other words, the algebraic sum of the potential differences around any closed path in a circuit must be zero. This law is based on the principle of conservation of energy.

Part(1)

The voltage Vx is calculated by applying Kirchhoff's voltage law in the first loop,

\(V_x - 0.5I_{B}(1)-4+0.5I_{B}(2)=0...................................(1)\)

Apply Kirchhoff's voltage law in the second loop,

\(-12+I_{B}(1)+0.5I_{B}(2) = 0........................(2)\)

From equations 1 and 2,

\(2I_{B} = 12\\I_{B} = 6A................................(3)\)

Substitute the value of \(I_B\\\) inequation (1),

Vx - 0.5(6) - 4 + 0.5(6) (2) = 0

Vx = 1 Volt

Part(2),

The power is calculated as,

\(P = 4 \times 0.5I_B\)

P = 4 x 0.5 x 6

P = 12 watts

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The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed (\(\omega\)), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel (\(v\)), in centimeters per second, is:

\(v = 2\pi\cdot r\cdot f\) (1)

Where:

If we know that \(f = 5.13\,hz\) and \(r = 41.5\,cm\), then the tangential speed of the chewing gum is:

\(v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)\)

\(v \approx 1337.659\,\frac{cm}{s}\)

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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The magnitude of OC is approximately 5.8 units, and the angle between the x-axis and vector OC is approximately 59.0 degrees

To evaluate the magnitude of vector OC, which is represented by vector C = (-3, -5), we can use the formula for the magnitude of a vector. The magnitude of a vector (denoted as ||v||) can be calculated using the Pythagorean theorem.

Magnitude of OC = ||C|| = √((-3)^2 + (-5)^2) = √(9 + 25) = √34 ≈ 5.8

To find the angle between the x-axis and vector OC, we can use trigonometry. The angle can be determined by finding the arctan of the y-component divided by the x-component of vector C.

Angle = arctan(-5 / -3) = arctan(5/3) ≈ 59.0 degrees

Therefore, the magnitude of OC is approximately 5.8 units, and the angle between the x-axis and vector OC is approximately 59.0 degrees (rounded to the nearest tenth).

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The magnitude of the force on the 1.0 nC charge is 3.8 N. The direction of the force on the 1.0 nC charge is towards the 2.0 nC charge, at an angle of 60°.

Magnitude is a measure of the size or strength of something. It can refer to physical quantities such as size, temperature, mass, or energy, as well as abstract qualities like loudness, brightness, or intensity. Magnitude is usually expressed as a number on a scale or as a unit of measurement, such as degrees or decibels. Magnitude can also refer to the importance or significance of an event or situation. For example, a hurricane may be said to have a magnitude of 5, or a financial crisis may have a magnitude of 10. Magnitude is an important concept in many scientific fields, such as physics and astronomy, where it is used to measure the size or power of an object or phenomenon.

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Answer: \(f=3.125 Hz\)

Explanation:

frequency = 1 / period

\(f=\frac{1}{T}\)

\(f=\frac{1}{0.32}=3.125Hz\)

Therefore, the frequency of the wave is 3.125 Hz.

Given:

The diameter of one salt granule is: d = 0.062 mm = 0.0062 cm.

The length of the paper is: l = 28 cm

The width of the paper is: b = 22 cm

To find:

A number of salt grains are needed to cover the paper of the given dimensions.

Explanation:

One sand granule when placed on the paper, will cover the area "a" which is given as:

The area "A" of the given paper can be calculated as:

Now, the number of salt grains "N" needed to cover the paper can be calculated as:

Final answer:

20403634 salt granules are required to cover the area of the given paper.

The length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

Given data: r = 0.2 mm = 0.2 x 10^-3m Resistivity = 3 x 10^-6 ohm R = 15.552 ohm

Formula Used: Resistivity (ρ) = (RA)/L

Where, R is resistance, A is the area of cross-section, L is the length of the wire.

Resistance (R) = ρ (L/A)

Multiplying A on both sides, we get

Resistance (R) x A = ρ L ... equation (1)

Area of the cross-section of a wire of radius (r) is given by, A = πr^2

where, π is a constant whose value is 3.14

Substituting the given values, we get

A = πr^2= π (0.2 x 10^-3m)^2= 1.2566 x 10^-7 m^2

Substituting the values of R, A and ρ in equation (1), we get

Length of wire (L) = (Resistance x Area) / Resistivity= (15.552 ohm x 1.2566 x 10^-7 m^2) / (3 x 10^-6 ohm)= 6.5268 m

Therefore, the length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

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