what is the frequency of a wave that travels 20 m/s with a wavelength of 200 meters?

The speed of the object after falling 5.0 m would be 5.4 m/s.

We can solve this problem using the law of conservation of energy, which states that the total energy of a system is constant. At the top of the drop, the object has potential energy equal to its mass times the acceleration due to gravity times its height above the ground:

Ep = mgh

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

At any point during the fall, the object has kinetic energy equal to one half its mass times its speed squared:

Ek = (1/2)mv^2

where v is the speed of the object.

Since there is no air resistance, the total energy of the system is conserved, so the initial potential energy at the top of the drop (Ep = mgh) is converted entirely into kinetic energy (Ek = (1/2)mv^2) as the object falls.

When the object has fallen 5.0 m, its potential energy is:

Ep = mgh = (2.0 kg)(9.8 N/kg)(5.0 m) = 98 J

The kinetic energy of the object at this point is equal to its initial potential energy minus the potential energy it still has at that point:

Setting the kinetic energy equation equal to the expression for Ek above and solving for v gives:

Therefore, the object's speed after falling 5.0 m is 5.4 m/s, rounded to 2 significant figures.

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Hitting the metal rod parallel to its axis will produce a longitudinal wave. By striking the rod perpendicular to its axis, you can produce a transverse wave.

By compressing the coils in a tiny area and then releasing the compressed area, a stretched spring can create a longitudinal wave. The disruption will then continue to spread as a longitudinal pulse.

We know that rarefaction happens in a longitudinal wave. Compression won't occur since a stretched string is inelastic and incompressible. Hence, a stretched string's longitudinal wave.

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It is challenging to provide a precise response to this query without knowing the skater's exact whereabouts. I can, however, give a general explanation of how potential energy, kinetic energy, mechanical energy, velocity, and height are related.

Kinetic energy (K.E.) plus potential energy equals mechanical energy (M.E. (P.E.)Kinetic Energy (K.E.) is equal to (1/2)mv.Potential Energy (P.E.) is defined as mgh.

The force acting on the two objects affects the potential energy formula. The formula for gravitational force is P.E. = mgh, where g is the acceleration caused by gravity (9.8 m/s2 at the earth's surface), and m is the mass in kilogrammes.

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A 1.24 kg bowling trophy is held at arm's length, a distance of 0.505 m from the shoulder joint. 6.14 Nm torque does the trophy exert about the shoulder if the arm is horizontal

To calculate the torque exerted by the 1.24 kg bowling trophy held at arm's length (0.505 m) from the shoulder joint when the arm is horizontal, you need to follow these steps:
Determine the force exerted by the trophy.
Since the trophy has a mass of 1.24 kg

The force exerted by the trophy due to gravity can be calculated using the equation:

F = m x g

where F is the force,

m is the mass, and

g is the acceleration due to gravity (approximately 9.81 m/[tex]s^2[/tex]).
F = 1.24 kg x 9.81 m/[tex]s^2[/tex] ≈ 12.16 N (Newtons)
Calculate the torque.
Torque (τ) is the rotational force that causes an object to rotate about an axis or pivot point. In this case, the axis is the shoulder joint.

The torque can be calculated using the equation:

τ = r x F x sin(θ)

where τ is the torque,

r is the distance from the axis to the point of force application (0.505 m),

F is the force exerted by the trophy (12.16 N), and

θ is the angle between the force vector and the distance vector.
Since the arm is held horizontally, the force exerted by the trophy is acting vertically downward, which means the angle θ between the force and distance vectors is 90 degrees.

The sine of 90 degrees is 1, so the equation simplifies to:

τ = r x F.
τ = 0.505 m x 12.16 N ≈ 6.14 Nm (Newton meters)

So, the torque exerted by the 1.24 kg bowling trophy about the shoulder joint when the arm is held horizontally at a distance of 0.505 m is approximately 6.14 Nm.
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If the aimless wanderer goes into a circular orbit 40,000 km above the surface of Mongo, it will take the ship approximately  11.81 hours to complete one orbit.

An orbit is a curved path followed by a celestial object around a star, planet, or moon, such as the path of the earth around the sun. Gravity keeps the object in orbit around the star or planet. An orbit can be circular or elliptical. Satellites, planets, moons, and asteroids all follow orbits around stars or planets.

The time required for the aimless wanderer to complete one orbit can be calculated by using the formula,
T = 2π√(a^3/GM)
Where:
a = altitude of orbit above the surface of the planet (radius of planet + altitude)
G = gravitational constant (6.67430 × 10^-11 N m^2/kg^2)
M = mass of the planet

First, we need to calculate the radius of the planet (R) based on the altitude of the orbit:
R = altitude of orbit + radius of planet
R = 40,000 km + radius of planet

We don't know the radius of the planet, so we can't calculate R directly. However, we can assume that the altitude of the orbit is small compared to the radius of the planet, which means we can approximate R as:
R ≈ 2 × altitude of orbit
R ≈ 2 × 40,000 km
R ≈ 80,000 km

Now that we have an approximation for the radius of the planet, we can calculate the period of the orbit:
T = 2π√(a^3/GM)
T = 2π√((R + altitude of orbit)^3/GM)
T = 2π√(((80,000 km + 40,000 km)^3)/(6.67430 × 10^-11 N m^2/kg^2 × M))

We don't know the mass of the planet, so we can't calculate T directly. However, we can make a simplifying assumption that the mass of the planet is much greater than the mass of the spacecraft, which means that the period of the orbit is only weakly dependent on the mass of the planet. Therefore, we can use the mass of Earth as an approximation for the mass of Mongo:
T ≈ 2π√(((80,000 km + 40,000 km)^3)/(6.67430 × 10^-11 N m^2/kg^2 × 5.9722 × 10^24 kg))
T ≈ 11.81 hours

Therefore, the spacecraft will take approximately 11.81 hours to complete one orbit of Mongo at an altitude of 40,000 km above its surface.

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The angular velocity of the rotating tires is 76.27 rad/s.

The angular velocity, ω, of a rotating object is the rate at which it rotates around a chosen center point,

ω = Δθ / Δt (angular displacement over time).

A truck with 0.347-m-radius tires travels at 26.4 m/s. We need to determine the angular velocity of the rotating tires in radians per second. The equation that relates speed and angular velocity is

v = ωr

where,ω = angular velocityv = linear velocity, r = radius of the tire. The angular velocity is given by the equation

ω = v / r

Substitute the given values into the formula;

ω = 26.4 m/s / 0.347 mω = 76.27 rad/s

Hence, the angular velocity of the rotating tires in radians per second of a truck with 0.347-m-radius tires that travels at 26.4 m/s is 76.27 rad/s.

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The time required for a spacecraft launched into a paraboliv trajectory at a perigee altitude of 200 km to leave the earth's sphere of influence is 7.77 days.

Thus, The terms apogee and perigee describe the separation between the Earth and the moon. The distance from the earth at apogee is the greatest.

The moon appears larger around perigee, which is when it is closest to the earth. Without anything to compare it against, the moon appears to be the same size when viewed from the ground. But the size disparity may actually be pretty substantial.

The tides on Earth are impacted by the moon's apogee and perigee. The moon has less gravitational attraction when it is at apogee, the distance from Earth at when it is closest to the moon.

Thus, The time required for a spacecraft launched into a paraboliv trajectory at a perigee altitude of 200 km to leave the earth's sphere of influence is 7.77 days.

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If pulley b is removed from the system, the ratio of the original value of the force with two pulleys to the new value of the force with one pulley is 2:1.

A pulley is a wheel with a groove for a rope or a belt to run over. Pulleys are commonly used to lift heavy loads. When the rope is looped around a pulley, it changes the direction of the force that must be exerted to lift the load. When you need to lift a load that is too heavy to lift alone, you can use a pulley system to reduce the amount of force required to lift the load.One of the most useful mechanical devices is a pulley system.

Pulleys are simple machines that can make it easier to lift heavy objects by changing the direction of the force required to lift them. Pulleys work by changing the direction of the applied force, making it easier to lift the load. The weight of the load is distributed over the multiple ropes of a pulley system so that the force required to lift the load is spread out over the multiple ropes, reducing the amount of force required to lift it.

*Complete question: If a pulley b is removed from a system, what is the ratio of the original value of the force with two pulleys to the new value of the force with one pulley?

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The increase in precipitation over an ocean leads to a decrease in the salinity of seawater.

An increase in precipitation over an ocean can result in a decrease in the salinity of the seawater. This is because the added freshwater from precipitation can dilute the seawater, lowering its salt concentration.

Additionally, increased precipitation can lead to increased runoff from land, which can carry freshwater and other dissolved substances into the ocean, further reducing the salinity of the seawater. However, the extent to which the salinity is affected will depend on various factors such as the rate and amount of precipitation, ocean currents, and evaporation rates.

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The radius of the copper cable is 0.107 mm.

The formula to calculate the radius of the cable is given by;

r=ρl/πV

Where, r is the radius of the cable

ρ is the resistivity of the copper

l is the length of the cable

V is the potential difference between two points on the cable.

The potential difference between two points on the cable is given by;

V=IR

Where, I is the current running through the cable

R is the resistance of the cable.

To determine the radius of the copper cable, we need to calculate its resistance first.

Resistance of the cable can be calculated as;

R=V/IR

Substitute the values given in the equation

R=0.016/(1200 A)=1.33x10^-5 Ω

Now, we can use this resistance value and resistivity of copper to calculate the radius of the cable.

The resistivity of copper is 1.72x10^-8 Ω.m.

So, r=ρl/π

[tex]V_r = 1.72x10^-^8 Ω.

m ×0.24 m/π ×1.33x10^-^5 Ω

r=0.000107 m = 0.107 mm[/tex]

So, the radius of the copper cable is 0.107 mm.

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Answer:

As the number of protons in the nucleus increases, the electronegativity or attraction will increase. Therefore electronegativity increases from left to right in a row in the periodic table

Explanation:

Answer:Fossil fuels are formed from the remains of plants and animals that lived millions of years ago. The amount of fossil fuels found in an area is influenced by several tectonic factors, including:

Depositional environment: The type of depositional environment where organic material was deposited affects the type and quantity of fossil fuels that are formed. For example, organic material deposited in marine environments may form oil, while organic material deposited in peat bogs or swamps may form coal.

Tectonic setting: Tectonic activity, such as plate movements, can affect the accumulation and preservation of organic material. For example, areas with active tectonic activity may have more uplift and erosion, which can result in the destruction of organic material before it can be buried and preserved.

Sedimentary basin characteristics: The size, depth, and shape of a sedimentary basin can also affect the amount of fossil fuels that are formed. For example, large, deep basins with thick layers of sedimentary rock may have a greater potential for oil and gas formation compared to smaller, shallower basins.

Source rock properties: The properties of the source rock, such as its organic content, mineral composition, and permeability, also influence the amount and type of fossil fuels that are formed. Rocks with higher organic content and better permeability are more likely to generate and retain larger amounts of fossil fuels.

Migration pathways: Once fossil fuels are formed, they must migrate through the surrounding rock to accumulate in economically viable amounts. The presence of faults, fractures, and other permeable structures can facilitate the migration and accumulation of fossil fuels in a particular area.

In summary, the amount of fossil fuels found in an area is affected by a combination of geological and tectonic factors that influence the formation, preservation, and accumulation of organic material over millions of years.

Explanation:

The components of acceleration are important and should not be neglected without introducing significant errors. are both the radial/centripetal and tangential.

The experiment in question is likely looking at the magnitude of acceleration in order to determine the force involved. The radial/centripetal component of acceleration is the change in direction of the velocity vector, and the tangential component is the change in magnitude of the velocity vector.

In the experiment, the radial/centripetal component of acceleration would likely be used to measure the direction of the force and the tangential component of acceleration would be used to measure the magnitude of the force. Therefore, if either one of these components were neglected, it would introduce error in the experiment.

Furthermore, depending on the type of experiment, the relationship between the two components of acceleration may be important in determining the force involved. For instance, a certain ratio between the two components of acceleration might suggest a particular type of force.

Therefore, neglecting either the radial/centripetal or the tangential component of acceleration would introduce significant errors in the experiment.

In conclusion, it is important to consider both the radial/centripetal and tangential components of acceleration in order to measure the force involved in an experiment. Neglecting either component could lead to inaccurate results and should therefore be avoided.

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Without knowing the details of the experiment or scenario in question, it is difficult to determine which component of acceleration (radial/centripetal or tangential) is more important.

In general, however, both components of acceleration can be important in certain situations. For example, in circular motion, both the radial/centripetal acceleration (directed towards the center of the circle) and the tangential acceleration (directed tangentially to the circle) contribute to the total acceleration of an object.

Neglecting one of these components of acceleration could introduce significant errors, depending on the specific situation. In some cases, neglecting the radial/centripetal acceleration could result in an object moving off its circular path, while neglecting the tangential acceleration could result in an object changing its speed or direction of motion.

Therefore, it is important to consider both components of acceleration in any situation involving circular motion, and to determine which component(s) are more significant for the specific scenario in question.

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Answer:

true

Explanation:

The apparent brightness of a star is how much energy is coming from the star per square meter per second, as measured on Earth. The further away the star is, the smaller the probability that a given photon emitted by the star will eventually hit Earth. Or said a quantitative form: all else being the same, the apparent brightness of a star is proportional the the inverse square of its distance.

Explanation:

P=40w

T=1hrs

=1×60=60

Now,

P2=40×60=2400

Hence, in a hour, the brain uses 2400w

Answer: Group 0 contains non-metal elements placed in the vertical column on the far right of the periodic table.

Explanation:

The elements in group 0 are called the noble gases. They exist as single atoms.

The work required to compress the spring an additional 1/2 m is also 10 joules. This is because the work done on the spring is equal to the potential energy stored in it.

When a spring is compressed or stretched, it stores potential energy. The amount of potential energy stored is proportional to the amount of deformation and the spring constant. The work required to compress or stretch the spring is equal to the potential energy stored in it. In this case, since it takes 10 newtons of force to hold the spring stretched 1/2 m from its equilibrium position, the potential energy stored in the spring is 10 joules. If the spring is compressed an additional 1/2 m, the work required is also 10 joules because the potential energy stored in the spring is still the same. This relationship between work and potential energy is a fundamental concept in the study of mechanics.

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The magnitude of the electric field between the plates is  [tex]1.19 \times 10^5 \ N/C[/tex], magnitude of the force on an electron between the plates is [tex]1.9 \times 10^{-14} N[/tex] and work done on the electron to move it to the negative plate is [tex]4.2 \times 10^{-17}\  J[/tex].

(a) The magnitude of the electric field between the plates can be calculated using the formula:

[tex]E = V/d[/tex]

where E is the electric field, V is the potential difference between the plates, and d is the distance between the plates.

Substituting the given values, we get:

[tex]E = 600 \ V / 5.04 \times 10^{-3} m = 1.19 \times 10^5 \ N/C[/tex]

Therefore, the magnitude of the electric field between the plates is [tex]1.19 \times 10^5 \ N/C[/tex].

(b) The magnitude of the force on an electron between the plates can be calculated using the formula:

F = qE

where F is the force, q is the charge of the electron ([tex]-1.6 \times 10^{-19} C[/tex]), and E is the electric field.

Substituting the given values, we get:

[tex]F = (-1.6 \times 10^{-19} C) \times (1.19 \times 10^5 \ N/C) = -1.9 \times 10^{-14}\  N[/tex]

Therefore, the magnitude of the force on an electron between the plates is [tex]1.9 \times 10^{-14} N[/tex].

(c) The work done on the electron to move it to the negative plate can be calculated using the formula:

W = qV

where W is the work done, q is the charge of the electron ([tex]-1.6 \times 10^{-19} C[/tex]), and V is the potential difference between the plates.

First, we need to calculate the electric potential at the initial position of the electron using the formula:

V = Ed

where E is the electric field and d is the distance between the electron and the positive plate.

Substituting the given values, we get:

[tex]V = (1.19 \times 10^5 N/C) \times (2.84 \times 10^{-3} m) = 338 \ V[/tex]

Therefore, the electric potential at the initial position of the electron is 338 V.

Substituting the given values, we get:

[tex]W = (-1.6 \times 10^{-19}\ C) \times (600\ V - 338\ V) = -4.2 \times 10^{-17}\  J[/tex]

Therefore, the work done on the electron to move it to the negative plate is [tex]4.2 \times 10^{-17}\  J[/tex]. Note that the negative sign indicates that the work is done by the electric field (i.e. the electron loses potential energy).

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By applying Newton's second law of motion the value of x component is 1 m/s^2 and the y component has the value of 0 m/s^2.

In contrast to the first law of motion, the second law of motion deals with the behaviour of objects when all external factors are in equilibrium.The second rule of motion, which is more quantitative, is frequently applied to determine what occurs when a force is present.
According to Newton's second rule, an object's acceleration is determined by its mass and the total force that is operating on it.The mass of the body has an inverse relationship with acceleration, which is exactly proportionate to the total force exerted on the body. This implies that as the force exerted on an object increases, so does the object's motion. Similar to how an object's motion decreases as its mass increases, so does its mass.
By applying Newton's second law of motion:-

for x component:

4N - 2N = 2a

2= 2a and a = 1 m/s^2.

for y component:

3N-(2N+1N)= 2a

a= 0m/s^2.

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The period and frequency of the motion is 1.33 second and 0.75 Hz respectively.

To solve this problem, we will use the formula for the period of simple harmonic motion. The period is the time taken for the object to complete one full oscillation. In this case, the object is the small cart on the frictionless track. We can use the following formula to calculate the period of the cart:

T=2L/v

where T is the period of the motion, L is the length of the track, and v is the speed of the cart.

In this case, L = 0.5 m and v = 0.75 m/s. Thus,

T=2(0.5 m)/(0.75 m/s)

T=1.33 s

The period of the motion is 1.33 seconds.

To find the frequency of the motion, we use the formula:

f=1/T

T=1.33 s

f=1/1.33

f=0.75 Hz

The frequency of the motion is 0.75 Hz.

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The formula for an electric field's work is as follows: The particle was pushed 1.25 metres in the electric field as a result of W = qEd.

When a charge is transported in an electric field, work is done. A positively charged particle, such a proton, would accelerate in the direction of the arrows if it were placed in an electric field.

W = qEd

Rearranging the formula to solve for d:

d = W/(qE)

Substituting the given values:

[tex]d = 15 J / (4 C * 3 N/C)[/tex]

d = 1.25 meters

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If you analyze the light from a low-density object such as a cloud of interstellar gas, you will see an emission spectrum.

An emission spectrum is a light spectrum created by the emission of light by a substance when it is exposed to high-frequency radiation. Each element has a unique line spectrum or emission spectrum, which can be used to detect the element.

An emission spectrum's line spectrum can be used to identify an element and distinguish it from other elements. It shows what color of light a substance emits when heated. The color of the lines on the emission spectrum is determined by the element that emitted the light. Each chemical element has a unique line spectrum, allowing astronomers to identify the elements in stars and other celestial objects.

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The block of timber weighs 0.72 grams in total.

The density of a material is defined as the mass of the material per unit volume. In this case, the density of the block of wood is given as 0.6 g/cm³, and the volume of the block is 1.2 cm³. To find the mass of the block, we can use the formula:

Mass = Density × Volume

Substituting the given values, we get:

Mass = 0.6 g/cm³ × 1.2 cm³

Simplifying the expression, we get:

Mass = 0.72 g

Therefore, the mass of the block of wood is 0.72 grams.

It is important to note that the units used in this calculation are consistent - the density is given in grams per cubic centimeter, and the volume is given in cubic centimeters. This ensures that the final answer for the mass is in grams, which is the appropriate unit for measuring the mass of a small object like a block of wood.

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The tangential speed of the roller coaster at the top of the loop is 8.64 m/s.

When a roller coaster travels through a loop, the normal force acting on the passengers changes to keep them moving in a circular path. At the top of the loop, the normal force is less than the weight of the passengers, while at the bottom of the loop, the normal force is greater than the weight of the passengers. In this case, we are given that the normal force is 3 times the weight of the passengers, which means that the normal force is: F₁ = 3mg, where m is the mass of the passengers and g is the acceleration due to gravity.

At the top of the loop, the normal force is equal to the centripetal force required to keep the passengers moving in a circle of radius r: F₁ = mv²/r, where v is the tangential speed of the roller coaster at the top of the loop. Combining these equations, we can solve for v:

mv²/r = 3mg

v² = 3gr

v = √(3gr)

Substituting the given values, we get: v = √(3 × 9.81 m/s² × 9 m) = 8.64m/s

So the tangential speed of the roller coaster at the top of the loop is 8.64m/s.

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Galaxies can exert a strong gravitational pull on each other when they are millions of light years from each other because gravity can reach huge distances.

Gravity is the force that maintains links to all bodies in the universe including celestial bodies such as planets and the sun or galaxies each other, which is required to avoid collision between them.

Therefore, with this data, we can see that gravity is able to maintain linked galaxies in the universe and thus they cannot collide

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The acceleration magnitude of block B when a cat of weight 45.8 N falls asleep on top of block A is 9.8 m/s².

Let the mass of block A be m₁ and its weight be W₁.

Let the mass of block B be m₂ and its weight be W₂.

The total mass that is resting on Block B is given by the equation:

m₁ + m₂ = 45.8/9.8

Where 9.8 m/s² is the acceleration due to gravity.

The net force acting on the block B is given by:

F = (m₁ + m₂)g

Where g is the acceleration due to gravity = 9.8 m/s²

The force exerted by block A on block B is given by:

F = m₁g

Therefore the net force on Block B is given by:

Fnet = (m₁ + m₂)g - m₁g

Fnet = m₂g

The acceleration of Block B is given by the equation:

Fnet = m₂a

Therefore, a = Fnet/m₂

We have, Fnet = m₂g

Therefore, a = g

Therefore, The acceleration of block B is equal to the acceleration due to gravity, g which is 9.8 m/s². Hence, the magnitude of its acceleration is 9.8 m/s².

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77.4 N of tension must be applied to the rope in order to create a 112 Hz wave's wavelength 1.135 m.

The distance that separates the crests and troughs of the light wave is known as the wavelength of light. With the Greek letter lambda (λ), it is identified. As a result, wavelength refers to the separation between one wave's peak or dip and the following wave.

The wave equation may be used to connect the cord's tension to the wave's wavelength and frequency:

v = fλ

where,

v denotes the wave's speed

The frequency is f, and

The wavelength is  λ

The elastic cord's elastic wave's velocity is provided by:

v = √(T/μ)

where,

T is the cord's tension, and

μ is the cord's linear mass density,

This is the mass per unit length:

μ = m/L

where m is the cord's mass and

Its length is L.

When we solve for the tension T using these formulas in the wave equation, we obtain:

T = μv² = μ(fλ)²

Inputting the values provided yields:

μ = m/L = 0.0204 kg / 4.21 m = 0.00484 kg/m

v = √(T/μ) --> v² = T/μ

λ = v/f --> v = λ*f

T = μv² = μ(fλ)²

= (0.00484 kg/m)*(1.135 m * 112 Hz)²

= 77.4 N

Hence, 77.4 N of tension must be applied to the rope in order to create a 112 Hz wave's wavelength 1.135 m.

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The strength of the magnetic field at the center of two concentric current loops is zero. The smaller loop has a radius of 0.0390 m and a current of 12.0 A. The larger current loop carries a current of 27.0 A, thus the radius of the larger loop should be 0.08775 m

What is the radius of the larger loop? Formula to calculate magnetic field on the axis of a circular current loop that is at a distance x from the center of the loop is given by the equation below.

B = [tex]\frac{\mu_oI}{2R}[/tex]

Here, B is the magnetic field on the axis of a circular current loop that is at a distance x from the center of the loop.

μ₀ is the permeability of free space.

I is the current in the loop,

R is the radius of the loop

In the case of a smaller loop, the magnetic field at its center is [tex] B_1 [/tex] and the magnetic field generated by the larger loop at the center of the smaller loop is [tex] B_2 [/tex].

Then [tex]B_1- B_2=0[/tex]

[tex]\frac{\mu_o}{2}[\frac{I_1}{R_1}-\frac{I_2}{R_2}]=0[/tex]

where [tex] I_1 [/tex] [tex] I_2 [/tex] are the currents in smaller and larger loops respectively.

[tex] R_1 [/tex] [tex] R_2 [/tex] are the radius of  smaller and larger loops respectively.

[tex]\frac{I_1}{R_1} = \frac{I_2}{R_2}\\R_2= \frac{I_2\times R_1}{I_1}\\R_2=0.08775 m.[/tex]

Therefore, the radius of the larger loop is 0.08775 m.

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The friction factor of the ice is M = 0.0102. The resistive friction force (Fr) defined either by ordinary or perpendicular force (N) pushing the particles close yields the coefficient of friction (fr), a numerical value.

The formula fr = Fr/N serves as a representation of it. By figuring out the greatest frictional force, someone can employ the frictional coefficient calculation to calculate the least force necessary to get an item going on a ground.

The item will advance whenever a force given to it that is larger than just the frictional force; as force is measured in newtons, as well as typical, which is a force, newtons are additionally utilized to measure normal. It's going to be cancelled out by the two military units. This coefficient of friction does not have a unit. It is a dimensionally and unitlessly empty quantity.

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