a formula that relatively accurately predicts the orbital distances of the first eight major bodies in the solar system and then somewhat predicts the next several is known as

The force that is most directly responsible for making current flow around the coil is an electric force.

An electric force is the force that arises between electrically charged objects, which includes protons, neutrons, and electrons. In order to drive an electric current through a wire, an electric field must be present. When a conductor, such as copper wire, is moved through a magnetic field, a current is generated inside the conductor in a phenomenon known as electromagnetic induction.

The current produced in the conductor is determined by the size of the magnetic field, the speed of the conductor, and the orientation of the magnetic field. An electric force is responsible for driving the current through the conductor, and the magnetic field is responsible for the creation of the current in the conductor. Therefore, it can be said that an electric force is most directly responsible for making current flow around the coil.

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The oscillatory behavior is a function of Φ (phi) instead of the period, T. Option A is correct answer.

Equation 9.4 describes the motion of a simple harmonic oscillator subjected to a periodic driving force. The term ? in the equation represents the frequency of the driving force, while in a simple harmonic oscillator, the oscillatory behavior is described by the period T. Therefore, the oscillatory behavior in Equation 9.4 is a function of frequency rather than the period. The amplitude Amod in Equation 9.4 is not twice the amplitude of a simple harmonic oscillator, nor is it time-dependent. Therefore, options B and C are incorrect. Option D is also incorrect since Equation 9.4 is different from the equation of a simple harmonic oscillator due to the presence of the driving force.

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Answer:9.81 m/s2.

Explanation:

If a 2 kg object is falling at 3 m/s, at what rate is gravity working on the object ? The object is falling at 3 m/s. Gravity is working on the object at a rate of 9.81 m/s2.

Answer:

Explanation:substitute the values for equation PE=m/vg×

The size of the force needed to keep the car of mass 1000kg moving around a circular track with radius 100m when the car is traveling at a constant speed of 36.4 m/s is 13456.16 N.

The size of the force needed to keep a car of mass 1000 kg moving around a circular track with radius 100 m when the car is traveling at a constant speed of 36.4 m/s is equal to the centripetal force. The centripetal force is the force which is directed towards the center of the circle and required to keep an object on a circular path.

Mathematically, the centripetal force (Fc) is equal to the mass of the car (m) times the square of its velocity (v2) divided by the radius (r) of the track.

Fc = mv²/r

Substituting the known values, we get:

Fc = 1000 kg * 36.42 m²/100 m = 13456.16 N

Therefore, the size of the force needed to keep the car moving in a circular track with the given conditions is 13456.16 N.

This force is an inward force, which is exerted by the track. This force keeps the car moving in a circular path and prevents it from veering off the track. It is equal to the product of the car's mass and the square of its velocity, divided by the radius of the track.

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The force can be calculated for an object with constant mass using the formula F = ma, where F is the force, m is the mass, and a is the acceleration. This formula is known as Newton's second law of motion.

It states that the force acting on an object is directly proportional to its mass and acceleration. Thus, the greater the mass of the object, the greater the force required to accelerate it at a given rate. For example, if a 2 kg object is accelerated at a rate of 5 m/s^2, the force required to achieve this would be F = 2 kg x 5 m/s^2 = 10 N. In summary, the force can be calculated for an object with constant mass by using Newton's second law of motion, which relates the force to the mass and acceleration of the object.

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To store 2.70x10^-3 J of electrical energy, capacitance c3 in the network must be 2.67μF. This can be calculated using the formula for energy stored in a capacitor network.

A network of capacitors is a collection of capacitors wired into a circuit. The capacitors store electrical energy as an electric field between their plates when a voltage is applied across the network. The formula E = 1/2 * C * V2 may be used to determine the total energy held in a capacitor network, where E is the energy held, C is the network's total capacitance, and V is the applied voltage. The capacitances of c1, c2, and c4 in this instance are specified as 4.00 F, 4.00 F, and 8.00 F, respectively. It is possible to determine that the capacitance c3 has to be 2.67 F to store 2.70 x 10-3 J of electrical energy by rearranging the formula and inserting the numbers.

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Answer:

To investigate the relationship between how far the spring is stretched and how much elastic potential energy is stored in the system, we can conduct the following experiment:

Materials:

Spring

Meter stick

Weights of different masses

Stopwatch

Triple beam balance

Retort stand

Clamp

Procedure:

Set up a retort stand and attach a clamp to it.

Attach the spring to the clamp.

Place a meter stick vertically next to the spring to measure the distance it is stretched.

Measure and record the natural length of the spring (when no weight is attached).

Add a known mass to the end of the spring and record the new length of the spring.

Calculate the difference between the stretched length and the natural length of the spring.

Repeat steps 5 and 6 for different masses.

Record the time taken for the spring to stretch when the mass is added.

Calculate the elastic potential energy stored in the spring for each mass added, using the formula:

Elastic potential energy = 1/2 x k x (distance stretched)^2

where k is the spring constant of the spring.

Plot a graph of the elastic potential energy stored against the distance stretched for each mass added.

Analyze the relationship between the distance stretched and the elastic potential energy stored.

By measuring the distance stretched and the time taken for the spring to stretch, we can calculate the velocity of the mass and use this to determine the elastic potential energy stored in the spring. By repeating this process for different masses and plotting a graph, we can determine if there is a linear relationship between the distance stretched and the elastic potential energy stored. This investigation will help to better understand the concept of elasticity and its relationship to potential energy.

The waterline separates the submerged section of the hull of the boat from the section above the water level.

The submerged section on the hull of a boat is separated from the section above the water level by the waterline. The waterline is the point where the surface of the water meets the hull of the boat. When a boat is placed in the water, the weight of the boat pushes down on the water and displaces it, creating a gap between the water and the hull. The waterline marks the level at which the boat displaces the water, separating the section of the hull that is below the waterline and submerged in water from the section above the waterline. The size and shape of the submerged section of the hull affect the boat's buoyancy, stability, and speed, making it an important design consideration for boatbuilders.

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Answer:

a) The phenomenon is known as the photoelectric effect.

b) When the zinc plate is exposed to UV light, some of the photons in the light have enough energy to knock electrons out of the surface of the plate. These emitted electrons carry a negative charge, so as they leave the surface of the plate, it becomes positively charged.

c) As the intensity of the UV light is increased, more photons with sufficient energy to knock electrons out of the surface of the plate are present. Therefore, the rate of emission of electrons increases.

d) The work function energy of zinc refers to the amount of energy required to remove an electron from the surface of a zinc atom. In other words, it is the minimum amount of energy required to cause the photoelectric effect to occur. The work function energy is a characteristic property of the material and is typically measured in electron volts (eV). In the case of zinc, the work function energy is 4.24 eV, meaning that at least 4.24 eV of energy is required to eject an electron from a zinc atom.

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The second proton must be 1.05 cm below the first proton to exert an electrostatic force equal in magnitude to the gravitational force between the two protons.

To find the distance between the two protons, we can set the electrostatic force equal to the gravitational force. The gravitational force on the first proton is given by Fg = mg, where m is the mass of the proton and g is the acceleration due to gravity near the surface of the Earth. The electrostatic force between the two protons is given by [tex]Fe = kq^2 / r^2[/tex], where k is Coulomb's constant, q is the charge of the proton, and r is the distance between the two protons.

Setting these two forces equal to each other, we get:

[tex]mg = kq^2 / r^2[/tex]

Solving for r, we get:

[tex]r = sqrt(kq^2 / mg)[/tex]

Substituting the values for the constants, we get:

[tex]r = sqrt((9 x 10^9 Nm^2/C^2) x (1.6 x 10^-19 C)^2 / ((1.67 x 10^-27 kg) x (9.81 m/s^2)))[/tex]

[tex]r = 2.17 x 10^-10 m[/tex]

Therefore, the second proton is located [tex]2.17 x 10^-10[/tex] meters directly below the first proton.

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The acceleration of an object with a mass of (m2-m1) is given by the force of F divided by the mass of the object, (m2-m1). This can be expressed as a = F/(m2-m1).


To solve this, first calculate the force F.

The force F acting on mass m1 is F1 = m1*a1 and the force F acting on mass m2 is F2 = m2*a2.

Since both masses are being acted upon by the same force, the equation F1 = F2 can be used.

Therefore, F = m1*a1 = m2*a2.

Now, substitute the known values for m1, m2 and a1, a2 into the equation to get F = m1*a1 = m2*a2.

F = m1*14.0 [tex]m/s^2[/tex] = m2*3.8 [tex]m/s^2[/tex].

Therefore, the acceleration of the object with a mass of (m2-m1) is given by a = F/(m2-m1). Substituting the value of F into the equation gives:

a = F/(m2-m1) = (m1*14.0 m/s2)/(m2-m1).

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The highway would be directed at an angle of 23.3 degrees with respect to due west.

To find the shortest length of highway between the two towns, we need to use the Pythagorean theorem. Let's draw a diagram:

  A

 /|

/ |

/  |

  |

  |   x

  |  /

  | /

  |/

  B

A represents the town that is 32.0 km south and 72.0 km west of town B. We can see that the distance between the two towns, x, is the hypotenuse of a right triangle with sides of length 32.0 km and 72.0 km.

Using the Pythagorean theorem, we can find x:

x = sqrt((32.0 km)^2 + (72.0 km)^2)

= 78.0 km

So the shortest length of highway that can be built between the two towns is 78.0 km.

Now, let's find the angle that this highway would be directed with respect to due west. We can use trigonometry for this. The angle we're looking for is the angle between the line connecting the two towns and due west.

  A

 /|

/ |

/  |  θ

  |

  |   x

  |  /

  | /

  |/

  B

We can see that tan(θ) = (32.0 km) / (72.0 km) = 0.444. Taking the inverse tangent of both sides, we get:

θ = [tex]tan^{-1}(0.444)[/tex]

= 23.3 degrees

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The minimum requirements for the design, materials, fabrication, erection, testing, and inspection of various types of piping systems are covered by ASME B31.3 Code.

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The option which represents a decrease in gravitational potential energy is B, where a girl jumps down from a bed.

Gravitational potential energy (GPE) is the energy a body has due to its position in a gravitational field. The higher an object is lifted, the greater its gravitational potential energy. The energy needed to lift an object comes from work done on it, which increases its potential energy by a comparable amount.

When the girl is on the bed, she is at a higher height than when she jumps down. Her height decreases from the bed, therefore the GPE will decrease. Thus, in the system where a girl jumps down from a bed, there is a decrease in gravitational potential energy.

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When light passes through a prism, the dispersion of light shows that different colors have different indices of refraction. When light passes through a medium, such as a prism, it is separated into its component colors, referred to as the visible spectrum.

This separation of light into its different colors is known as the dispersion of light. The index of refraction is the measure of how much a ray of light bends when it passes through a medium. When light passes through a medium with a high index of refraction, it bends more than when it passes through a medium with a low index of refraction.

When light passes through a prism, it bends or refracts due to the differences in the index of refraction of each color of light.  Because each color of light has a slightly different index of refraction, they bend at different angles, causing the light to spread out and separate into its component colors, as seen in a rainbow.

The dispersion of light when it passes through a prism shows that different colors have different indices of refraction.

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The angular velocity of the turntable is 4.19 radians/second . The angular accelatation of the turntable is 8.38 radians/second².

The torque on the grammaphone record is 0.0419 Nm. The rotational kinetic energy is 0.0439 J. . The work done by the torque in 0.2 s is 0.00702 J.

The turntable has a uniform angular velocity of 40 rounds per minute. To convert this to radians per second, we can use the following conversion factors:

1 round = 2π radians (since there are 2π radians in a complete circle)

1 minute = 60 seconds

So, the angular velocity (ω) can be calculated as:

ω = 40 rounds/minute × (2π radians/round) × (1 minute/60 seconds)

ω ≈ 4.19 radians/second

The record accelerates uniformly to the angular velocity of the turntable in 0.5 seconds. Let's denote the angular acceleration as α. We can use the formula:

ω = ω₀ + αt

Plugging in the values, we get:

4.19 = 0 + α(0.5)

α = 4.19 / 0.5

α ≈ 8.38 radians/second²

The moment of inertia of the grammaphone record (I) is given as 5.0 x 10^(-3) kgm². We can use the formula:

τ = Iα

Plugging in the values, we get:

τ = (5.0 x 10^(-3)) x 8.38

τ ≈ 0.0419 Nm

The rotational kinetic energy (K) can be calculated using the formula:

K = 0.5 * I x ω²

Plugging in the values, we get:

K = 0.5 * (5.0 x 10^(-3)) * (4.19)²

K ≈ 0.0439 J

The work done (W) by the torque can be calculated using the formula:

W = τ * θ

Plugging in the values for α and t = 0.2s, we get:

θ ≈ 0.5 * 8.38 * (0.2)²

θ ≈ 0.1676 radians

Now, we can calculate the work done:

W = 0.0419 * 0.1676

W ≈ 0.00702 J

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A child sleds down a steep, snow-covered hill with an acceleration of 2.82m/s squared. if her initial speed is 0.0 m/s and her final speed is 15.5 m/s, It takes the child 5.50 seconds to travel from the top of the hill to the bottom.

We can use the following kinematic equation to solve this problem:

v^2 = u^2 + 2as

Where:

v is the final speed

u is the initial speed

a is the acceleration

s is the distance traveled

We are given u = 0.0 m/s, a = 2.82 m/s^2, and v = 15.5 m/s. We need to find s.

Rearranging the equation gives:

s = (v^2 - u^2) / (2a)

Substituting the values gives:

s = (15.5^2 - 0.0^2) / (2 x 2.82) = 74.47 m

Now, we can use another kinematic equation to find the time taken:

v = u + at

Where t is the time taken.

Substituting the values gives:

15.5 = 0.0 + 2.82t

Solving for t gives:

t = 15.5 / 2.82 = 5.50 seconds

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A ball revolving on the end of a thin string in a circle of radius 1.25 m at an angular speed of 10.4 rad/s angular momentum of the ball revolving on the end of the string is 3.058 m²/s.

The angular momentum (L) of an object revolving in a circle is given by the formula:

[tex]\[ L = I \cdot \omega \][/tex]

Here:

I = moment of inertia of the object about the axis of rotation.

[tex]\(\omega\)[/tex] = angular speed of the object in radians per second.

For a point mass rotating about an axis at a distance r from the axis, the moment of inertia (I) is given by:

[tex]\[ I = m \cdot r^2 \][/tex]

Here:

m = mass of the object.

r = radius of the circle.

Given the values:

Mass (m) = 0.210 kg

Radius (r) = 1.25 m

Angular speed ([tex]\(\omega\)[/tex]) = 10.4 rad/s

Let's calculate the moment of inertia (I) first:

[tex]\[ I = m \cdot r^2 \\\\= (0.210 \, \text{kg}) \cdot (1.25 \, \text{m})^2 \][/tex]

Now, calculate the angular momentum using the moment of inertia and angular speed:

[tex]\[ L = I \cdot \omega \][/tex]

Plug in the values and calculate:

[tex]\[ L = (0.210 \, \text{kg} \cdot (1.25 \, \text{m})^2) \cdot (10.4 \, \text{rad/s}) \][/tex]

Calculate the angular momentum (L):

[tex]\[ L \approx 3.058 \, \text{kg} \cdot \text{m}^2/\text{s} \][/tex]

Thus, the angular momentum of the ball revolving on the end of the string is approximately [tex]\(3.058 \, \text{kg} \cdot \text{m}^2/\text{s}\)[/tex].

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When Kate stops oscillating and finally comes to rest, she will be hanging 1.96 meters below the bridge.

This is because, the time period T of a pendulum depends on its length l and acceleration due to gravity g. Mathematically, T = 2π √(l/g)

Therefore, l = (gT²)/(4π²)

Given, T = 4s and g = 9.8m/s²

We can find the length of the pendulum. Hence, l = (9.8m/s²×(4s)²)/(4π²) = 1.96m

Therefore, Kate's distance below the bridge will be 1.96 meters.

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The radio waves that WPL emits have a wavelength of roughly 3.14 metres.

An FM radio wave's frequency and wave size must be known in order to determine its wavelength. An FM radio wave has a frequency of 94.1 MHz. FM radio waves are measured in metres. An FM radio wave therefore has a wavelength of [tex]94.1*106[/tex] metres.

c = fλ

where [tex]c = 3.00 x 10^8 m/s[/tex] (speed of light in a vacuum).

We need to convert the frequency from MHz to Hz:

[tex]95.5 MHz = 95.5 x 10^6 Hz[/tex]

Now we can solve for the wavelength:

c = fλ

λ = c/f

[tex]λ = (3.00 x 10^8 m/s) / (95.5 x 10^6 Hz)[/tex]

λ = 3.14 m

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When falling, raindrops often develop electric charges, which is accurate. However, these charges are normally quite small, thus they are not likely to create significant forces between the droplets.

What are raindrops? Raindrops are drops of water that fall from the atmosphere, as they grow heavier due to gravity. As they fall, they may accumulate electric charges.

When drops are charged, they can attract one another or repel one another, depending on their relative charges.

What are electric charges? The electric charge is the fundamental property of matter that distinguishes it from other properties such as mass or energy. Objects that have opposite charges attract each other, whereas objects that have the same charge repel each other.

Suppose two 1.8 mg drops each have a charge of 29 pC, and the centers of the droplets are at the same height and 0.44 cm apart. We want to determine whether this creates noticeable forces between the droplets. We can use Coulomb's law to calculate the electric force between the two drops:

[tex]F_{electric} = \frac{kq_1q_2}{r^2}[/tex]

where [tex]k = 9\times10^9 N m^2/C^2[/tex]

k is Coulomb's constant,

q1 and q2 are the charges of the droplets, and

r is the distance between their centers. T

he masses of the droplets aren't essential for this calculation, as they do not affect the electric force much.

From Coulomb's law, F =  1.35 x [tex]10^{-16}[/tex] N.

This force is incredibly small, as expected.

Therefore, it is unlikely that there would be noticeable forces between the droplets due to their electric charges when falling through the air.

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Given Information:

[tex]m_{1}=1.3 \ kg[/tex] (Mass of the block attached to the pendulum)

[tex]l=1.7 \ m[/tex] (Length of the string)

[tex]m_{b}=0.01 \ kg[/tex] (Mass of the bullet)

Information we want to Find:

[tex]||\vec v_{b} || = ?? \ m/s[/tex]

Concepts/Equations used:

Using the idea of momentum, momentum conservation, and energy conservation to solve.

Momentum => [tex]\vec p=m \vec v[/tex]

Momentum Conservation => [tex]\vec p_{0} = \vec p_{f}[/tex]

Energy Conservation => [tex]E_{0} = E_{f}[/tex]

Finding the blocks velocity at the bottom of its swing using conservation of energy. Analyzing points 1 and 2 (Refer to the attached image).

Energy at point 1...

The energy at point 1 is all gravitational potential energy, where [tex]U_{g} =m_{1} gl[/tex].

[tex]\Longrightarrow U_{g} =m_{1} gl[/tex]

[tex]\Longrightarrow U_{g} =(1.3)(9.8)(1.7)[/tex]

[tex]\Longrightarrow U_{g} =(1.3)(9.8)(1.7)[/tex]

[tex]\Longrightarrow U_{g} = 21.658 \ J[/tex]

Thus, [tex]E_{0}= 21.658 \ J[/tex].

Now for the energy at point 2...

The energy at point 1 is all kinetic, where [tex]K=\frac{1}{2}m_{1}v^{2} _{f}[/tex].

[tex]\Longrightarrow K=\frac{1}{2}m_{1}v^{2} _{f}[/tex]

[tex]\Longrightarrow K=\frac{1}{2}(1.3)v^{2} _{f}[/tex]

[tex]\Longrightarrow K=0.65v^{2} _{f}[/tex]

Thus, [tex]E_{f}=0.65v^{2} _{f}[/tex].

[tex]E_{0} = E_{f}[/tex]

[tex]\Longrightarrow 21.658 = 0.65v^{2} _{f}[/tex]

[tex]\Longrightarrow v^{2} _{f}=33.32[/tex]

[tex]\Longrightarrow v _{f}=\sqrt{33.32}[/tex]

[tex]\Longrightarrow v_{f} = 5.77 \ m/s[/tex]

The momentum of the block at point 2, [tex]\vec p =m_{1} \vec v_{f}[/tex].

[tex]\vec p_{block} =m_{1} \vec v_{f}[/tex]

[tex]\Longrightarrow \vec p_{block} =(1.3) (5.77)[/tex]

[tex]\Longrightarrow \vec p_{block} = 7.50 \ Ns[/tex]

For the block to stop the momentum of the bullet must equal the momentum of the moving block.

[tex]\vec p_{0} = \vec p_{f} = > \vec p_{bullet} = \vec p_{block}[/tex]

[tex]\Longrightarrow m_{b} \vec v_{b} = 7.5[/tex]

[tex]\Longrightarrow (0.01) \vec v_{b} = 7.5[/tex]

[tex]\Longrightarrow \vec v_{b} = 750 \ m/s[/tex]

Thus, the bullet was travelling 750m/s before hitting the block.

To maintain a constant data read rate, the linear velocity of the DVD must remain constant. Since the circumference of the disk increases as the radius increases, the rotational speed must also increase as we move from the center to the outer edge. The linear velocity is given by:

v = ωr

where v is the linear velocity, ω is the rotational speed in radians per second, and r is the distance from the center of the disk.

Since we want to maintain a constant linear velocity, we have:

v = constant

ω1r1 = ω2r2

where ω1 is the rotational speed at the center of the disk, r1 is the radius of the center of the disk, ω2 is the rotational speed at the outer edge of the disk, and r2 is the radius of the outer edge of the disk.

We know that the disk spins at 1600 rpm for information at the center of the DVD. Let's assume that the radius of the center of the disk is 1 cm, and the radius of the outer edge of the disk is 6 cm. Then we have:

ω1 = 1600 rpm = 167.55 rad/s

r1 = 1 cm

r2 = 6 cm

Substituting these values into the equation above, we can solve for ω2:

ω1r1 = ω2r2

167.55 x 1 = ω2 x 6

ω2 = 27.92 rad/s

Finally, we can convert the angular velocity to rpm:

ω2 = 27.92 rad/s x 60 s/min ÷ 2π rad = 266.08 rpm

Therefore, the rotational speed of the disk needs to be 266.08 rpm at the outer edge of the DVD to maintain a constant data read rate.

One can use the kinetic theory of gases to explain why the pressure of gas increases as the temperature rises in the following ways:At the molecular level, gases are made up of a large number of small particles that are in constant motion.

The movement of the gas particles produces kinetic energy. The total kinetic energy of the particles in a gas is proportional to the temperature of the gas.When the temperature of a gas is increased, the kinetic energy of its particles increases as well.

As the particles move faster, they collide more frequently and with greater force. The force exerted by the particles on the walls of the container is the pressure of the gas.In conclusion, as the temperature of the gas increases, the average kinetic energy of the gas particles increases, which in turn leads to more collisions between particles and with the walls of the container. As a result, the pressure of the gas increases.

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The work done by the electric force to move a 1 C charge from point A to B is 5 J.

The work done by an electric force to move a 1 coulomb (1 c) charge from point A to point B is calculated using the formula: Work = Force x Distance. As the electric force is equal to the charge multiplied by the electric field (F = qE), the work done is calculated as: Work = (1 c) x (Electric Field) x (Distance from A to B). The result is given in Joules (J). The work done by the electric force to move a 1 C charge from point A to B is calculated by multiplying the charge and the potential difference between the two points.
Therefore, the work done is given by: Work done = qΔV Where:q = 1 CΔV = VB - VA. The electric potential difference between point A and B can be obtained from the electric potential at each point. Therefore, the electric potential difference is given by: ΔV = VB - VA = 10 - 5 = 5 V. Substituting the values of q and ΔV in the above equation, we get:Work done = qΔV = 1 x 5 = 5 J.

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(a) Same-direction currents result in attraction. Force per unit length: 1.11 × 10⁻⁵ N/m.

(b) Opposite direction currents result in repulsion. Force per unit length: -1.11 × 10⁻⁵ N/m.

When two parallel wires carry currents in the same direction, they experience a force of attraction, while they experience a force of repulsion when the currents are in opposite directions. The force per unit length exerted on one of the wires by the other can be calculated using the formula F = μ₀I₁I₂L / (2πd), where F is the force per unit length, μ₀ is the permeability of free space, I₁ and I₂ are the currents in the wires, L is the length of the wires, and d is the distance between the wires. For the given problem, the force per unit length exerted on one of the wires by the other is 1.11 × 10⁻⁵ N/m when the currents are in the same direction, indicating attraction, and -1.11 × 10⁻⁵ N/m when the currents are in opposite directions, indicating repulsion. Therefore, the wires either attract or repel each other based on the direction of the currents they carry.

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So, in Activity 2-2, if the coil is pulled away from the north pole of a magnet, it is likely that an EMF will be induced in the coil.

However, based on general principles of electromagnetism, if a coil is pulled away from the north pole of a magnet, it will experience a change in magnetic flux, which can induce an electromotive force (EMF) in the coil. This phenomenon is known as electromagnetic induction.

The magnitude and direction of the induced EMF will depend on several factors, such as the strength of the magnet, the velocity at which the coil is moved, and the orientation of the coil with respect to the magnetic field. If the coil is part of a circuit, the induced EMF can cause a current to flow in the circuit.

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The statement about dark matter that is false is "it extends to 200,000ly from the center of the galaxy."

Dark matter refers to the invisible matter in space that is thought to be present in galaxies in large amounts. It cannot be seen, felt, or heard, but its gravitational effects can be observed.

The following statements about dark matter are true:

It may be undiscovered subatomic particles.

It makes up more mass compared to luminous matter.

It makes up about 80% of the total mass of our galaxy.

It is in a halo that is nearly spherical.

Dark matter does not extend to 200,000ly from the center of the galaxy. Dark matter is spread throughout the galaxy and beyond, extending far beyond the visible disk of the galaxy.

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The total work done by the 40.0-kg child pushing the wagon 1.50 m up a ramp at an angle of 14.0 degrees with the horizontal is 106.6 J. This includes the work done both on the body and the crate.

To solve this problem, we must first calculate the work done on the body of the child. This can be done using the equation W = F×d×cos θ where F is the force exerted by the child, d is the distance the wagon is pushed, and θ is the angle of the ramp.

In this case, F = 175 N, d = 1.50 m, and θ = 14.0°. So, the work done on the body is:

W = 175 N × 1.50 m × cos 14.0°

= 67.9 J

Now, we must calculate the work done on the crate. This can be done using the equation W = F×d×sin θ, where F, d, and θ are the same values as above.

So, the work done on the crate is:

W = 175 N × 1.50 m × sin 14.0°

= 38.7 J

Adding the work done on the body and the crate, the total work done by the child is 67.9 J + 38.7 J = 106.6 J.

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