A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnitude, the balls will be attracted to each other.

a. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.

b. The negative charges on the sphere A move to the side closer to sphere B, while the positive charges move to the side further away from sphere B. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.

c. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

d. The negative charges on the sphere B move to the side closer to sphere A, while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

Answer:

Electric force will be equal to [tex]2\times 10^{-4}N[/tex]

Explanation:

We have given charge [tex]q=2\mu C=2\times 10^{-6}C[/tex]

Electric field will be [tex]E=100N/C[/tex]

We have to find the magnitude of force

Electric force is the multiplication of electric field and charge

So electric force [tex]F=qE=2\times 10^{-6}\times 100=2\times 10^{-4}N[/tex]

Electric force will be equal to [tex]2\times 10^{-4}N[/tex]

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency [tex]f=7.0\times10^{13}\ Hz[/tex]

We need to calculate the reduced mass

Using formula of reduced mass

[tex]\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}[/tex]

Where, [tex]m_{H}[/tex]= atomic mass of H

[tex]m_{I}[/tex]= atomic mass of I

Put the value into the formula

[tex]\mu=\dfrac{1\times126.9}{1+126.9}[/tex]

[tex]\mu=0.99\ u[/tex]

[tex]\mu=0.99\times1.66\times10^{-27}\ Kg[/tex]

[tex]\mu=1.643\times10^{-27}\ kg[/tex]

We need to calculate the force constant of the spring

Using formula of frequency

[tex]f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}[/tex]

[tex]k=f^2\times 4\pi^2\times\mu[/tex]

Put the value into the formula

[tex]k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}[/tex]

[tex]k=317.8\ N/m[/tex]

Hence, The force constant of the spring is 317.8 N/m.

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

hope it helped <3

Answer:

R=12 m

Explanation:

I have drawn a vector form which I attached here.Please first go through that

Given Data

y (Tree Taller) = 9.0 m

x(Distance from center to edge is)= 8.5 m

To find

R(magnitude of the butterfly's displacement

)

Solution

[tex]R=\sqrt{x^{2} +y^{2} }\\R=\sqrt{9^{2}+8.5^{2}  }\\ R=12.379m[/tex]

Convert to two significant figures

R=12 m

The butterfly's displacement, when it moves from the tree to the flower, is approximately 19m. This is determined using the Pythagorean theorem, which is applied to the '9m' height difference between the tree and flower, and '17m' width of the garden.

To calculate the magnitude of the butterfly's displacement, we need to use the Pythagorean theorem, which establishes a relationship in any right triangle. The vertical height of the tree is 9m, which forms one side of the right triangle. The horizontal distance across the garden (or width of the garden) is 17m, forming the other side of the right triangle.

Applying the Pythagorean theorem (a^2 + b^2 = c^2), where 'a' and 'b' are the sides and 'c' is the hypotenuse (in this case, the butterfly's displacement), we get:

Displacement = √[(9m)^2 + (17m)^2]

So, the displacement of the butterfly is approximately 19 m, when rounded to two significant figures.

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Answer:

2.92 x 10¹² electrons

Explanation:

given,

distance between two plastic ball, r = 17 cm

                                      r = 0.17 m

Force of attraction = F = 68 mN

                F = 0.068 N

number of electron transferred from one ball to another.

using Coulomb Force equation

[tex]F = \dfrac{kq^2}{r^2}[/tex]

[tex]0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}[/tex]

q² = 2.1835 x 10⁻¹³

q = 4.67 x 10⁻⁷ C

now, number of electron

 [tex]N = \dfrac{q}{e}[/tex]

e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C

  [tex]N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}[/tex]

       N = 2.92 x 10¹² electrons

electrons were transferred from one ball to the other is 2.92 x 10¹²

Field strength is directly proportional to the square of charge. The charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex]

[tex]F = \dfrac {kQ^2}{r^2}[/tex]

Where,

[tex]F[/tex]- field strength = 68 mN = 0.06 N

[tex]Q[/tex]- charge

[tex]r[/tex]- distance =  17 cm =0.17 m

[tex]k[/tex]- constant = 9x10⁹

Put the values in the formula,

[tex]0.068 = \dfrac{9\times 10^9 \rimes Q^2}{0.017^2}\\\\q^2 = 2.1835 \times 10^{-13}\\\\q = 4.67 \times 10^{-7} \rm \ C[/tex]

Therefore, the charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex].

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A) Intensity: [tex]4.8 W/m^2[/tex]

B) Intensity: [tex]2.1 W/m^2[/tex]

C) Total intensity: [tex]13.8 W/m^2[/tex]

D) Distance: [tex]1.38\cdot 10^5 m[/tex]

Explanation:

A)

The relationship between intensity of a sound and power emitted is given by

[tex]I=\frac{P}{A}[/tex]

where

I is the intensity

P is the power

A is the surface area at the distance considered

For the speaker in this problem:

P = 60 W

The distance is r = 1.0 m, so the area to consider is the surface of a sphere with this radius:

[tex]A=4\pi r^2 = 4\pi(1.0)^2=12.6 m^2[/tex]

Therefore, the intensity is

[tex]I=\frac{60}{12.6}=4.8 W/m^2[/tex]

B)

For this problem, we can use the same equation we used before:

[tex]I=\frac{P}{A}[/tex]

where in this case, we have:

P = 60 W is the power of the speaker

r = 1.5 m is the distance considered

Therefore, the area of the spherical surface is

[tex]A=4\pi r^2 = 4\pi(1.5)^2=28.3 m^2[/tex]

And so, the intensity here is

[tex]I=\frac{60}{28.3}=2.1 W/m^2[/tex]

C)

In this problem, we have to consider 4 speakers. In order to find the total intensity, we have to add the intensity of each speakers.

For the two front speakers, we have

[tex]P=60 W[/tex]

r = 1.0 m

So we have already calculated their intensity in part A), and it is

[tex]I_1 = I_2 = 4.8 W/m^2[/tex]

For the two rear speakers, we have

[tex]P=60 W[/tex]

r = 1.5 m

So their intensity has been calculated in part B),

[tex]I_3 = I_4 = 2.1 W/m^2[/tex]

Therefore, the total intensity is

[tex]I=I_1 + I_2 + I_3 + I_4 = 4.8+4.8+2.1+2.1=13.8 W/m^2[/tex]

D)

In this case, we want to find the distance r at which the intensity is

[tex]I=10^{-12} W/m^2[/tex]

Using four speakers with power

P = 0.06 W

Since we have 4 speakers, the intensity due to each speaker at the location considered must be

[tex]I'=\frac{I}{4}=\frac{10^{-12}}{4}=2.5\cdot 10^{-13} W/m^2[/tex]

Using the usual equation, we find the area:

[tex]A=\frac{P}{I'}=\frac{0.06}{2.5\cdot 10^{-13}}=2.4\cdot 10^{11} m^2[/tex]

And so, we can find what is the corresponding distance:

[tex]A=4\pi r^2\\r=\sqrt{\frac{A}{4\pi}}=\sqrt{\frac{2.4\cdot 10^{11}}{4\pi}}=1.38\cdot 10^5 m[/tex]

So, the person must be at 138 km.

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The intensity of sound waves can be computed by using the equation I = P / (4πr²). For the minimum discernible intensity, we employ a rearranged version of the same formula to calculate the distance from the car stereo at the threshold of hearing.

The question is about the calculation of sound intensity produced by a car stereo featuring four speakers, each rated at 60 Watts, at various distances. To calculate the intensity (I) of the sound waves at different distances, we use the equation I = P / (4πr²) where P represents power and r is the distance.

For part D, to find the distance (d) at which sound from the stereo could still be discerned, we rearrange the intensity equation to solve for distance, d = √(P / (4πI)). We substitute P with the sound produced (0.06 W) and I with the minimum discernible intensity (10-12 W/m²). Disclaimer: Since external factors like absorption or diffusion are not taken into account, and the car stereo is not treated as a perfect point source emitter, these values serve as approximation.

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Answer:

1008.33 W.

Explanation:

Power: This can be defined as the rate at which energy is consumed or dissipated. The S.I unit of power is Watt (W).

P = E/t.......................... Equation 1

Where P = Energy, E = Energy, t = time.

But from the question,

E = 1/2mΔv²................... Equation 2

where m = mass of the sprinter, Δv = change in velocity of the sprinter = final velocity - initial velocity.

Substitute equation 2 into equation 1

P = 1/2mΔv/t....................... Equation 3

Given: m = 50 kg, Δv = 11-0 = 11 m/s, t = 3.0 s.

Substitute into equation  3

P = 1/2(50)(11)²/3

P = 25(121)/3

P = 1008.33 W.

Thus the power output = 1008.33 W.

Final answer:

The power output of a 50-kg sprinter who accelerates from 0 to 11 m/s in 3.0 s is 1008.33 Watts or approximately 1.35 horsepower.

Explanation:

To calculate the power output of a 50-kg sprinter who accelerates from 0 to 11 m/s in 3.0 s, we use the work-energy principle and the definition of power. The work done on the sprinter is equal to the change in kinetic energy, and power is the work done per unit time.

The kinetic energy (KE) at the start is 0 since the sprinter starts from rest. The kinetic energy at the end is KE = 1/2 x mass x velocity² = 1/2 x 50 kg x (11 m/s)². After calculating the kinetic energy, we get KE = 1/2 x 50 x 121 = 3025 Joules.

Now, we find the power by dividing the work by the time interval: Power = Work / Time = 3025 J / 3.0 s = 1008.33 Watts.

To convert watts to horsepower, use the conversion 1 horsepower = 746 watts. Therefore, Power in horsepower = 1008.33 W / 746 W/hp = approximately 1.35 hp.

Answer:

156.67 m/s

0.45676 times the speed of sound

Explanation:

Distance from the ground = 23.5 km = 23500 m

Time taken by the blast waves to reach the ground = [tex]2\ minutes\ 30\ seconds=2\times 60+30=150\ s[/tex]

Spedd of the wave would be

[tex]Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s[/tex]

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

[tex]\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v[/tex]

The blast wave is 0.45676 times the speed of sound

The average velocity of the blast wave can be calculated using the distance traveled and time taken, resulting in 156.7 m/s. This velocity is less than half the speed of sound at sea level.

The average velocity of the blast wave can be calculated by dividing the distance traveled by the time taken. The distance is the altitude where the explosion occurred, 23.5 km. Converting 2 minutes 30 seconds to seconds gives a time of 150 seconds.

So, velocity = distance / time = 23.5 km / 150 s = 0.1567 km/s = 156.7 m/s.Comparing this to the speed of sound at sea level, which is 343 m/s, the blast wave's velocity of 156.7 m/s is less than half the speed of sound.

Answer:

The magnitude of the magnetic field vector is 1.91T and is directed towards the east.

The steps to the solution can be found in the attachment below.

Explanation:

For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.

Fm = Fg

qvBSin(theta) = mg

Where q = magnitude of charge

v = magnitude of the velocity vector = 4 x10^4 m/s

B = magnitude of the magnetic field vector

theta = the angle between the magnetic field and velocity vectors = 90°

m = mass of the charge = 0.195g

g = acceleration due to gravity =9.8m/s²

On substituting the respective values of all variables in the equation (1) above

B = 1.91T

The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.

Below is the step by step procedure to the solution.

Final answer:

The magnitude of the minimum magnetic field that will keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction can be calculated using the equation for magnetic force. We need to find the minimum value of B that produces a force greater than or equal to the gravitational force acting on the particle. By setting up the equation qvBsinθ ≥ mg and solving for B, we can find the minimum magnetic field strength required.

Explanation:

The magnitude of the minimum magnetic field that will keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction can be calculated using the equation for magnetic force. The force on a moving charged particle in a magnetic field is given by the equation F = qvBsinθ, where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. In this case, we can assume that the angle between the velocity and the magnetic field is 90 degrees, as the particle is moving horizontally northward. Therefore, to find the minimum magnetic field that keeps the particle moving in this direction, we need to find the minimum value of B that produces a force greater than or equal to the gravitational force acting on the particle.

Let's take the example given in Example 22.1, where a glass rod with a positive charge of 20 nC is thrown with a horizontal velocity of 10 m/s due west in a place where the Earth's magnetic field is due north parallel to the ground. The force on the rod due to the Earth's magnetic field can be calculated using the equation F = qvBsinθ, where q is the charge of the rod (20 nC), v is the velocity of the rod (10 m/s), B is the magnetic field strength (unknown), and θ is the angle between the velocity and the magnetic field (90 degrees). The force on the rod should be equal to or greater than the gravitational force acting on it to keep it moving in the same horizontal, northward direction. Therefore, we can set up the equation F ≥ mg, where F is the force on the rod, m is the mass of the rod, and g is the acceleration due to gravity. Substituting the values, we get qvBsinθ ≥ mg. Solving for B, we get B ≥ mg / qvsinθ. Plugging in the given values, we can find the minimum magnetic field strength required to keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction.

Answer:

b

Explanation:

Answer:

B

Explanation:

egde test 2023

Answer

given,

weight = 1000 lbf

g = 32.174 ft/s²

mass =[tex] \dfrac{1000}{32.174}[/tex]

m = 31 slug

1 slug = 14.593 kg

m = 31 x 14.593

m = 452.383 Kg

b) weight on moon

  W = m g

   W = 452.383 x 1.62

   W= 732.86 N

c) we know,

   F = m  a

  400 lbf = 31 slugs x a

  a = 12.90 ft/s²

1 ft = 0.304 m

  a = 3.92 m/s²

Final answer:

This answer covers calculating mass in kg, weight on the moon, and acceleration with a force on both Earth and the moon.

Explanation:

Mass Calculation:

Weight on Moon:

Acceleration with Force:

Answer:

300000 N

Explanation:

We are given that

Mass of jumbo jet=300,000kg

Acceleration of jet=[tex]4m/s^2[/tex]

We have to find the force generated by each of its four engines.

We know that

Force,F =ma

Total force exert by four engines=4F

[tex]4F=300000\times 4[/tex]

[tex]F=\frac{300000\times 4}{4}=300000N[/tex]

Hence, the force generated by each of its four engines=300000 N

Final answer:

To determine the actual rate of heat input to the geothermal power plant, calculate the heat input using the provided formula. The actual thermal efficiency can be found by dividing the net power output by the heat input. The actual rate of heat rejection can then be calculated as the difference between the heat input and the net power output. To determine the power output if the geothermal water leaves the plant at a different temperature while maintaining its thermal efficiency, use the provided formula with the appropriate values.

Explanation:

The actual rate of heat input to the geothermal power plant can be calculated using the formula:

Heat input = mass flow rate * specific heat capacity * (hot water temperature - environment temperature)

Given that the mass flow rate is 210 kg/s, the specific heat capacity is 4.186 J/g°C, the hot water temperature is 150°C, and the environment temperature is 25°C, we can substitute these values into the formula to find the heat input.

The actual thermal efficiency of the power plant can be calculated using the formula:

Thermal efficiency = (Net power output / Heat input) * 100%

Given that the net power output is 8000 kW and we have already calculated the heat input, we can substitute these values into the formula to find the actual thermal efficiency. The maximum possible thermal efficiency can be calculated as the Carnot efficiency, which is the maximum possible efficiency for a heat engine operating between two temperatures.

The actual rate of heat rejection from the power plant can be calculated as the difference between the heat input and the net power output.

To find the power output if the geothermal water leaves the plant at 40°C and maintain its thermal efficiency, we can use the formula:

Power output = (Heat input - Heat rejection) * (Net power output / Heat input)

Substituting the appropriate values into the formula will give us the desired power output.

To develop this problem we will apply the relationship of density, such as the unit of mass per unit volume of a body. For this relationship we will use the known constants of the value of the land's mass and its respective radius. These values will be converted from kilograms to grams and meters to centimeters respectively. We will find the volume through the geometric relationship of the sphere using the radius of the earth.

The mass of the Earth is given as

[tex]m_E = 5.9722*10^{24}kg (\frac{1000g}{1kg}) = 5.9722*10^{27}g[/tex]

The radius of the Earth is

[tex]r_E = 6.3781*10^6m (\frac{100cm}{1m}) = 6.3781*10^8cm[/tex]

Using the geometric value of volume for a sphere, and using the radius of the earth, as the radius of that sphere we have to

[tex]V_e = \frac{4}{3} \pi r^3[/tex]

Replacing,

[tex]V_E = \frac{4}{3} \pi (6.3781*10^8)^3[/tex]

[tex]V_E = 1.09*10^{27}cm^3[/tex]

The expression for the density of the Earth is

[tex]\rho = \frac{m_E}{V_E}[/tex]

Replacing,

[tex]\rho = \frac{5.9722*10^{27}}{1.09*10^{27}}[/tex]

[tex]\rho = 5.48g/cm^3[/tex]

Therefore the average density of the earth is [tex] 5.48g/cm^3[/tex]

Answer:

The following are true:

A satellite's motion is independent of its mass.

The launch speed of a satellite determines the shape of its orbit around Earth.

Explanation:

A satellite's motion is independent of its mass.

The speed of a satellite can be determined by means of the Universal law of gravity:

[tex]F = G\frac{M\cdot m}{r^{2}}[/tex] (1)

Where G is the gravitational constant, M and m are masses of the two objects and r is the distance between them.

In equation 1, Newton's second law can be replaced:

[tex]F = m\cdot a[/tex]

[tex]m. a = G\frac{M\cdot m}{r^{2}}[/tex]   (2)

Since it is a circular motion, the centripetal acceleration can be used:

[tex]a = \frac{v^{2}}{r}[/tex] (3)

Then, equation 3 can be replaced in equation 2:

[tex]m\frac{v^{2}}{r} = G\frac{M\cdot m}{r^{2}}[/tex]  (4)

In this case, m is the satellite's mass and M is the Earth mass. Therefore, v can be isolated from equation 4:

[tex]v^{2} = G\frac{M\cdot m r}{mr^{2}}[/tex]

[tex]v^{2} =\frac{G M}{r}[/tex]

[tex]v = \sqrt{\frac{G M}{r}}[/tex]

Where G is the gravitational constant, M is the mass of the Earth and r is the orbital radius of the satellite.

Notice, how the speed of the satellite does not depend in its mass, only in the Earth mass and its orbital radius.

The launch speed of a satellite determines the shape of its orbit around Earth.

Depending on the launch speed the satellite can reach a lower or higher height above the surface of the Earth and, therefore, its orbital radius will be bigger or smaller according with that height.

Remember that the orbital radius for the satellite will be the sum of the Earth radius and the height above the surface.      

Key term:

Geosynchronous satellite: It is a satellite with the same orbital period as Earth rotation period (24 hours).  

Answer:

84%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

[tex]y_0=1\ m[/tex]

x = 10 m

t = Time taken

[tex]v_0[/tex] = 3.5 m/s (assumed, as it is not given)

[tex]A_0=\pi 1.5^2[/tex]

We have the equation

[tex]y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s[/tex]

[tex]x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s[/tex]

From continuity equation we have

[tex]Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2[/tex]

Fraction is given by

[tex]f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%\approx 84\ \%[/tex]

The fraction is 84%

A fraction of the cross-sectional area of the hose hole that Isabella must cover is 84%.

Given the following data:

Vertical distance = 1.0 meters.

Horizontal distance = 10.0 meters.

Radius = 1.5 cm.

Vertical speed = 3.5 m/s.

The time taken to reach a maximum height is given by this formula:

[tex]H=\frac{1}{2} gt^2\\\\t=\sqrt{\frac{2H}{g} } \\\\t=\sqrt{\frac{2\times 1.0}{9.8} }[/tex]

t = 0.452 seconds.

For the velocity required, we have:

[tex]x=x_o+Vt\\\\10=0+V0.452\\\\V=\frac{10}{0.452}[/tex]

V = 22.12 m/s.

From continuity equation, we have:

[tex]A=\frac{A_0V_0}{V} \\\\A=\frac{\pi r^2V_0}{V}\\\\A=\frac{\pi1.5^2 \times 3.5}{22.12} \\\\A=\frac{24.7433}{22.12}[/tex]

A = 1.118 cm^2.

Now, an expression for the fraction of the cross-sectional area is given by:

[tex]F=\frac{A_0-A}{A_0} \times 100\\\\F=\frac{7.0695-1.118}{7.0695} \times 100\\\\F=\frac{5.9515}{7.0695} \times 100\\\\F=0.8419 \times 100[/tex]

F = 84.19 ≈ 84%.

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Answer:

The maximum potential energy of the system is 0.2 J

Explanation:

Hi there!

When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.

The equation of elastic potential energy (EPE) is the following:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretching distance.

The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.

Then:

EPE = 1/2 · 40 N/m · (0.1 m)²

EPE = 0.2 J

The maximum potential energy of the system is 0.2 J

The maximum potential energy of the spring system, given a spring constant of 40 N/m and a displacement of 0.1 m, is calculated using the formula PEmax = (1/2)kx2, and the result is 0.2 J.

When dealing with a spring system in physics, we use Hooke's Law, which states the force exerted by a spring is directly proportional to the amount it is stretched or compressed and is given by F = -kx, where k is the spring constant and x is the displacement from equilibrium. The potential energy stored in such a system can be calculated using the formula PEmax = (1/2)kx2.

In this case, given a spring constant k = 40 N/m and a maximum displacement x = 0.1 m, the potential energy at maximum compression or stretch is:

PEmax = (1/2)kx2 = (1/2)(40 N/m)(0.1 m)2 = (1/2)(40)(0.01) = 0.2 J.

Therefore, the maximum potential energy of the oscillating system is 0.2 J.

Answer:

[tex]d=2.5367\times 10^{-6}\ cm[/tex]

Explanation:

Given:

Now the distance moved by the tectonic plate in the stipulated times:

[tex]d=v.t[/tex]

[tex]d=\frac{1}{365\times 12\times 3600} \times 80[/tex]

[tex]d=2.5367\times 10^{-6}\ cm[/tex]

Tectonic plates are the mass rock from of the earth's uppermost mantle and the crust which show similar behavior of movement with respect to the other such groups of lithosphere.