What Enlightenment idea is represented by the headline

a) $$E_1 = \frac{(9.0 \times 10⁹ N m²/C²)(2.6 \times 10⁻⁶C)}{(3.0 m)²} \approx 7.80 \times 10⁵ N/C$$, direction is to the right ; b) $$E_2 = \frac{(9.0 \times 10⁹ N m²/C²)(1.4 \times 10⁻⁶ C)}{(4.0 m)²} \approx 3.94 \times 10⁵ N/C$$, electric field is directed towards point charge so, direction is to the left  c) $$|\vec{E}| = \√{E_1² + E_2²} \approx 8.86 \times 10⁵ N/C$$ and its direction is up.

Charge that exists in a body that has fewer electrons than protons is known as positive electrons.

a. To find the electric field at the origin due to charge Q1, we can use the formula for the electric field due to point charge:

$$E_1 = \frac{k Q_1}{r_1²}$$

k is Coulomb constant (k = 9.0 × 10⁹ N m²/C²), Q1 is the charge, and r1 is the distance from the charge to the point where we want to find the electric field.

Q1 = 2.6 μC and r1 = 3.0 m (since x1 = -3.0 m is the distance from Q1 to the origin).

$$E_1 = \frac{(9.0 \times 10⁹ N m^2/C²)(2.6 \times 10⁻⁶C)}{(3.0 m)²} \approx 7.80 \times 10⁵ N/C$$

The electric field is directed away from point charge, so direction of the electric field at the origin due to Q1 is to the right (positive x direction).

b. Similarly, to find the electric field at the origin due to charge Q2, we use the same formula:

$$E_2 = \frac{k Q_2}{r_2²}$$

where Q2 = 1.4 μC and r2 = 4.0 m (since x2 = 4.0 m is the distance from Q2 to the origin).

$$E_2 = \frac{(9.0 \times 10⁹ N m²/C²)(1.4 \times 10⁻⁶ C)}{(4.0 m)²} \approx 3.94 \times 10⁵ N/C$$

The electric field is directed towards point charge, so direction of the electric field at the origin due to Q2 is to the left.

c. $$\vec{E} = \vec{E_1} + \vec{E_2}$$

$\vec{E_1}$ is the electric field due to Q1 and $\vec{E_2}$ is the electric field due to Q2.

net electric field at the origin is: $$|\vec{E}| = \√{E_1² + E_2²} \approx 8.86 \times 10⁵ N/C$$ and its direction is up.

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According to computer models, planetary migration is most likely to occur in a planetary system early in its history, when there is still a gaseous disk around the star.

Planetary migration is the process by which a planet changes its orbital position over time. The process is often caused by gravitational interactions with other planets or a planetesimal disk, which causes the planet to migrate inward or outward from its original orbit.

Other factors that can contribute to planetary migration include the late stages of a star's evolution when a stellar wind clears the gaseous disk away and asteroids and comets occasionally collide with planets.

However, early in a planetary system's history, when there is still a gaseous disk around the star, is the most likely time for planetary migration to occur.

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To determine the total power delivered to the circuit (i.e., the total power dissipated in the resistors), you can use the formula:

P = I²R ; where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.

To find the current, you can use Ohm's law:

V = IR

where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms.

Here's an example:

Suppose you have a circuit with two resistors, R1 and R2, connected in series.

The voltage across the circuit is 10 volts, and the resistances of the two resistors are 2 ohms and 4 ohms, respectively. You can find the total resistance of the circuit by adding the resistances of the two resistors:

R = R1 + R2 = 2 + 4 = 6 ohms

To find the current in the circuit, you can use Ohm's law:

I = V/R = 10/6 = 1.67 amps

Then, you can find the power dissipated in each resistor:

P1 = I²R1 = (1.67)²(2) = 5.56 wattsP2 = I²R2 = (1.67)²(4) = 11.11 watts

And finally, you can find the total power dissipated in the circuit by adding the power dissipated in each resistor:Ptotal = P1 + P2 = 5.56 + 11.11 = 16.67 watts

So the total power delivered to the circuit is 16.67 watts.

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The final kinetic energy of the block is 308.98 J.

Let's solve the problem using the work-energy theorem.

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy

W = ΔKE

Initially, the block is at rest. Therefore, its initial kinetic energy is zero.

Ki = 0

We have to find the final kinetic energy of the block. Hence, Kf = ?

Work done on the block

W = Fscosθ

Work done by the applied force,

F = 75 Ns = 8 mθ = 37°

W = Fscosθ

W = 75 × 8 × cos 37°

W = 451.27 J

Work done by the frictional force

Ff = μkFn

The normal force

Fn = mg

Fn = 6 × 9.8

Fn = 58.8 N

Here,

Ff = μkFn

Ff = 0.28 × 58.8

Ff = 16.51 J

Work of friction:

W = Ff × s

W = 16.51 × 8

W = 132.1 J

The total work done on the block,

Wtotal = W + Wfriction

Wtotal = 451.27 + 132.1

Wtotal = 583.37 J

According to the work-energy theorem,

Wtotal = ΔKE

ΔKE = Wtotal

ΔKE = 583.37 J

Final kinetic energy of the block

Kf = KEFinal

Kf = ΔKE

Kf = 583.37 J

Kf = 308.98 J

Therefore, the final kinetic energy of the block is 308.98 J.

Complete question:

A 6 kg block is pushed 8m up a rough 37 degree inclined plane by a horizontal force of 75 N. The initial speed of the block is 2.2 m/s up the plane and a constant kinetic friction force of 25 N opposes the motion. Calculate the fianl kinetic energy of the block.

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The given statement "an object floating in a container of water and partially submerged has the same density as the water" is true.

When an object is placed in water, it sinks until the weight of the water displaced by the object equals the weight of the object.

If an object has the same density as water, it displaces an equal amount of water to its own weight. When it displaces the same amount of water that has an equivalent mass to the object, it will float partially submerged. If the object's density is greater than water, it will sink. If the object's density is less than that of water, it will float entirely above the water's surface.

Density is defined as the mass of an object per unit volume. The formula for density is mass/volume. Density is a crucial physical property that is used to define and classify materials. The density of an object is determined by its mass and volume. The unit of measurement for density is kg/m3 or g/cm3. The density of water is 1 g/cm3, which is why objects with a density of less than 1 g/cm3 float on water.

An object floating in a container of water and partially submerged has the same density as the water.

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The frequency of a standing wave with a wave speed of 12 m/s as it travels on a 4.0-m string fixed at both ends is 3.0 Hz.

A standing wave is produced by a wave with the same amplitude, frequency, and wavelength moving in the opposite direction with the initial wave. This indicates that the wave appears to stand in one place. Standing waves can only be generated in a medium if there is a boundary that restricts the movement of the wave. Standing waves can be observed in various shapes and sizes, and their frequencies are determined by a variety of factors, including the wave speed and the length of the string. When a standing wave is generated in a string, the points where the wave appears to be fixed are known as nodes, while the points where the string vibrates with the most amplitude are known as antinodes.In this scenario, the wave speed and the length of the string are given.

The wave speed, frequency, and wavelength of a wave are related by the formula v = fλ, where v is the wave speed, f is the frequency, and λ is the wavelength. Since the length of the string is fixed, the wavelength of the standing wave is twice the length of the string. Thus, λ = 2L = 8 m. Plugging in the values for the wave speed and wavelength, the frequency can be calculated as follows:f = v / λ = 12 m/s / 8 m = 1.5 Hz. The frequency of a standing wave with a wave speed of 12 m/s as it travels on a 4.0-m string fixed at both ends is 3 Hz.

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The electric field stored between two square plates of 9.3 cm on a side and separated by a 2.5 mm air gap is 1110 N/C. This can be calculated using Coulomb's law and the given information.

Coulomb's law states that the electric field is equal to the charge (Q) divided by the permittivity of free space (ε₀) multiplied by the distance (d) squared:

E=Q/(ε₀*d²).
Plugging in the given information,

E=(13 nC)/(8.85 x 10⁻¹² * 0.0025²) = 1110 N/C.
This answer uses Coulomb's law to calculate the electric field stored between two square plates, given the plates' side lengths, air gap width, and charge magnitude.

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Blood flows with a speed of 30 cm/s along a horizontal tube with a cross-section diameter of 1.6 cm.The speed of blood flow in the part of the same tube that has a diameter of 0.8 cm is 15 cm/s.

To arrive at this answer, we can use the formula for the flow rate of a fluid in a pipe:
Q = A × V
where Q is the flow rate, A is the cross-sectional area of the pipe, and V is the velocity of the fluid.
Therefore, if we substitute the values for A and V of the first section, we can calculate the flow rate for that section:
Q1 = A1 × V1
Q1 = π ×(1.6 cm/2)² × 30 cm/s
Q1 = 24.72 cm³/s
Now we can use the flow rate and the cross-sectional area of the second section to calculate the velocity of the fluid:
Q1 = A2 × V2
V2 = Q1 / A2
V2 = 24.72 cm³/s / (π × (0.8 cm/2)²)
V2 = 15 cm/s
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The mass, in units of me (the mass of the earth), of a planet with twice the radius of the earth for which the escape speed is twice that of the earth is 8 me.

The amount of matter in an object is referred to as mass. Mass is expressed in terms of the unit kilogram in the International System of Units (SI).

The escape velocity is defined as the minimum velocity required for an object to leave the gravitational influence of another object. For example, if a ball is thrown from the surface of the earth at a speed of 11.2 km/s (40,320 km/h), it will escape the earth's gravitational pull and continue into space.

The formula for escape velocity is given by:

  v=√(2GM/r)

Where, v is the escape velocity, G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet.

The formula for mass:

  m = v²r/Gm = (2v)²(2r)/GMm = 8r/G

Therefore, the mass, in units of me (the mass of the earth), of a planet with twice the radius of earth for which the escape speed is twice that of the earth is 8 me.

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(a) The equivalent capacitance of the 4.20 µF and 8.50 µF capacitors when connected in series is approximately 4.2017 µF.

(b) The equivalent capacitance of the 4.20 µF and 8.50 µF capacitors when connected in parallel is 12.70 µF.

When two capacitors are connected in series, the equivalent capacitance is given by the formula,

1/Ceq = 1/C1 + 1/C2

where C1 and C2 are the capacitances of the two capacitors.

Substituting the given values,

1/Ceq = 1/4.20 µF + 1/8.50 µF

1/Ceq = 0.238 µF^-1

Ceq = 1 / (0.238 µF^-1)

Ceq = 4.2017 µF (rounded to four significant figures)

When two capacitors are connected in parallel, the equivalent capacitance is given by the formula,

Ceq = C1 + C2

where C1 and C2 are the capacitances of the two capacitors.

Substituting the given values,

Ceq = 4.20 µF + 8.50 µF

Ceq = 12.70 µF

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The primary mirror of an astronomical telescope is often shaped and polished to a parabolic shape because a parabolic shape allows for the mirror to collect the most amount of light and focus the parallel rays of light to a single point for better image clarity.

The reason that the primary mirror of an astronomical telescope is often shaped and polished to a parabolic shape is to reduce spherical aberration.

What is an astronomical telescope?An astronomical telescope is an optical instrument that aids in the observation of remote objects by collecting electromagnetic radiation such as visible light. It consists of two primary components: a primary mirror or lens that gathers and focuses light, and an eyepiece or camera that magnifies and projects the image formed by the primary.

A parabolic shape is a mirror or lens that has a curve that is more curved in the center than at the edges, and it is often used in astronomical telescopes to reduce spherical aberration. Spherical aberration is an optical defect that causes the image of a point source to become fuzzy and blurred. It occurs when the rays passing through the edges of a spherical lens or mirror become focused at a different distance than those passing through the center. This causes the image to be blurred around the edges, which makes it difficult to view small or distant objects. Parabolic mirrors are used to correct this problem because they are designed to focus all incoming light to a single point, resulting in a sharper and clearer image.

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The intensity of the electromagnetic radiation from the Sun just outside the atmosphere of the Earth is 1.55 x 10-9 W/m2.

The intensity of electromagnetic waves from the sun just outside the atmosphere of the Earth can be calculated using the inverse-square law.

This law states that the intensity of the radiation decreases with the square of the distance from the source. Thus, the intensity of the radiation at the edge of the atmosphere will be lower than that at the surface of the Sun.

The intensity of the radiation, we need to know the distance from the Sun to the Earth. This distance is approximately 93 million miles (150 million kilometers).

The intensity of the radiation at the edge of the atmosphere by taking the inverse-square of this distance, which is approximately 1.55 x 10-9 W/m2.

This is the intensity of the electromagnetic radiation from the Sun just outside the atmosphere of the Earth.

The intensity of the electromagnetic radiation from the Sun just outside the atmosphere of the Earth is 1.55 x 10-9 W/m2.

This is due to the inverse-square law, which states that the intensity of radiation decreases with the square of the distance from the source.

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The magnitude of the torque on the current loop is calculated using the formula τ=BIA sinθ, where B is the magnitude of the magnetic field, I is the current, A is the area of the loop, and θ is the angle between the magnetic field and the loop's plane. In this case, the magnitude of the torque is τ = (1.2 T)(0.5 A)(5 cm x 5 cm)sin(30°) = 7.5 x 10-3 Nm.

The torque is the rotational force that causes the loop to rotate. This is due to the fact that a force is exerted on the loop by the magnetic field when there is a current running through it. This force generates a torque on the loop, which will cause it to rotate until the angle between the plane of the loop and the magnetic field is 0°.

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The angular velocity assuming a constant speed for a track-star who runs 400 m circular track in 45 s is 0.139 radians/s.

To calculate the angular velocity first the circumference of the track is 400 meters.

This means that the angular displacement of the track star during the race is:

θ = s / r

where θ is the angular displacement,

s is the distance traveled by the track star (which is equal to the circumference of the track), and

r is the radius of the circular track.

2.) Since the radius of the circular track is half of its diameter, we have:

r = 400 m / 2 = 200 m

Plugging this into the equation for angular displacement, we get:

θ = 400 m / 200 m = 2π radians

3.) Next, we can use the formula for angular velocity:

ω = θ / t

where ω is the angular velocity and

t is the time it takes for the track star to complete the race.

4.)Plugging in the values we have:

ω = θ / t

ω = 2π radians / 45 s

Therefore, the angular velocity of the track star is:

ω = 0.139 radians/s (rounded to three significant figures)

Therefore, the track star's angular velocity assuming a constant speed is approximately 0.139 radians/s

The angular displacement of the track star is equal to one complete revolution around the circular track, which is equal to 2π radians.

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Final velocity of Jeff's car is 7.133 m/s south. The direction is 59.3° south of east.

In this issue, we can utilize preservation of energy to track down the last speed and course of Jeff's crash mobile after the impact with Julia's. Before the impact, the energy in the x-heading is zero, and in the y-course, it is 60 kg × 4 m/s = 240 kg⋅m/s north. Julia's force is 45 kg × 6 m/s = 270 kg⋅m/s west.After the crash, the energy in the x-course is rationed. The absolute energy in the x-course is as yet zero, as Julia's force that way is likewise zero. In the y-heading, the absolute force after the crash is 60 kg × vj + 45 kg × 2 m/s sin 15°, where vj is Jeff's last speed in the y-course.Utilizing protection of energy, we can compare the force when the crash in the y-heading:

60 kg × 4 m/s + 45 kg × 6 m/s = 60 kg × vj + 45 kg × 2 m/s sin 15°

Working on this situation, we get:

240 kg⋅m/s + 270 kg⋅m/s = 60 kg × vj + 12.19 kg⋅m/s

Addressing for vj, we get:

vj = (240 kg⋅m/s + 270 kg⋅m/s - 12.19 kg⋅m/s)/60 kg

vj = 7.133 m/s south

Consequently, Jeff's last speed is 7.133 m/s south. To find the course, we can utilize geometry. The point of Jeff's last speed concerning the x-pivot is given by:

θ = tan^-1(vj/4 m/s)

θ = 59.3° south of east

Accordingly, the last speed and heading of Jeff's amusement cart are 7.133 m/s at a point of 59.3° south of east.

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The time it takes the object to fall through the change in speed using the impulse-momentum theorem is 0.62 seconds.

The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse exerted on it.

To calculate the time it takes the object to increase it speed using the  impulse-momentum theorem, we use the formula below.

Formula:

Recall that F/m = acceleration. Therefore,

Where:

From the question,

Given:

Substitute these values into equation 1 and solve for t

Hence, the time it takes the object to fall is 0.62 seconds.

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The current that flows through the 200 Ω resistor is 1.56 A.

Given resistance values of 200 Ω, 250 Ω, and 1000 Ω are connected in parallel across a source. The source current is 6 A. We are required to find the current that flows through the 200 Ω resistor.

Recall that when resistors are connected in parallel, the current is divided among them. And the voltage across each resistor is the same. The equivalent resistance of three parallel resistors is given by;

1/Rp = 1/R1 + 1/R2 + 1/R3Rp = (R1 x R2 x R3)/(R1R2 + R1R3 + R2R3)

Put the values into the formula;

Rp = (200 x 250 x 1000)/(200×250 + 200×1000 + 250×1000)

Rp = 52.17 Ω

The total current in the circuit, It = 6 A

From Ohm's Law;

V = IR,

where V is the voltage across each resistor

V1 = V2 = V3V = I×R

Therefore; V = I×Rp

The current flowing through the 200 Ω resistor, I1 = V1/200 = I × Rp/200The current flowing through the 200 Ω resistor, I1 = (6×52.17)/200I1 = 1.56 A

Thus, the current that flows through the 200 Ω resistor is 1.56 A.

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True. An incandescent bulb may have a higher initial cost than other types of lightbulbs, but it uses less energy over its lifetime and thus reduces energy costs.

For an incandescent bulb, the given statement is true. In candescent bulbs are traditional bulbs, which use a filament to create light. These bulbs are less efficient, as they waste most of the electricity they use as heat rather than light. As a result, the bulbs are less cost-effective in the long run.

They use up more energy than modern alternatives such as CFLs (compact fluorescent lights) or LEDs (light-emitting diodes). Despite their low initial cost, incandescent bulbs are not recommended for long-term use. They consume more electricity and thus have a greater impact on the environment. Therefore, it is not true that the energy costs of an incandescent bulb will be low over its life time.

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Time taken for the boat to make a complete round trip of 1 100 m upstream followed by a 1 100-m trip downstream is 200 seconds.

The boat moves at 10.8 m/s relative to the water, and the current is 2.00 m/s. To make a complete round trip of 1 100 m upstream followed by a 1 100-m trip downstream, it would take:

When the boat is moving upstream, it is going against the direction of the current.

Upstream: 1 100 m/ (10.8 m/s - 2.00 m/s) = 102.78 s

When the boat is moving downstream, it is going in the same direction as the current,

Downstream: 1 100 m/ (10.8 m/s + 2.00 m/s) = 97.22 s

Total time taken in going upstream and downstream is the sum of the time calculated in both cases

102.78 s + 97.22 s = 200 s

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The waves will cancel each other out and no waves will remain. If two waves of the same frequency, but different amplitudes, interfere with each other, the resulting wave will have an amplitude equal to the sum of the two wave amplitudes.

Pulse waves are pressure waves that are created as the heart pumps blood throughout the body. They are detected through pulse points, such as on the wrists, neck, or temples. Pulse waves can be measured using a device called a pulse oximeter, which uses a sensor to detect the pressure of the pulse wave.

Pulse waves can provide information about a person’s heart rate and oxygen saturation levels.

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The process described in the question is known as scenario planning. It is a strategic planning method that involves generating multiple plausible scenarios of future conditions and analyzing the potential impact of each scenario on an organization or a system.

Scenario planning is a useful tool for decision-making, risk management, and identifying opportunities in an uncertain or rapidly changing environment.

By developing a range of scenarios, decision-makers can anticipate potential challenges and opportunities and develop strategies to respond effectively to each situation.

This approach allows organizations to be better prepared and more resilient in the face of future uncertainties. Scenario planning can be applied to various fields, including business, economics, environmental planning, and public policy.

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If a star is 11 pc away from us, its apparent visual magnitude will be lower than its absolute visual magnitude. The star's apparent magnitude would be only 0.38 magnitudes lower than its absolute magnitude.

This is because the apparent magnitude of a star is affected by its distance from us. As the distance increases, the star appears dimmer, and its apparent magnitude decreases.

The distance modulus formula gives us a way to calculate the difference between the apparent and absolute magnitudes of a star:

Distance modulus = 5 * log(distance in parsecs) - 5

For a star that is 11 pc away, the distance modulus is,

Distance modulus = 5 * log(11) - 5 = 1.38

This means that the star's apparent magnitude will be 1.38 magnitudes lower than its absolute magnitude.

If the same star were only 5 pc away from us, the distance modulus would be,

Distance modulus = 5 * log(5) - 5 = 0.38

In this case, the star's apparent magnitude would be only 0.38 magnitudes lower than its absolute magnitude. This means that the star would appear brighter and have a higher apparent magnitude when it is closer to us.

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The sound level for this noise is approximately 50 decibels.

Sound level is a logarithmic measure of the ratio between the sound pressure level of a particular sound wave and a reference level. The reference level is typically set at the threshold of human hearing, which corresponds to an intensity of 10^-12 W/m^2. The sound level (measured in decibels, dB) of a sound wave is given by,

L = 10 log10(I/I0)

where I is the intensity of the sound wave and I0 is the reference intensity, which is typically set at 10^-12 W/m^2.

So, for an intensity of 10^-7 W/m^2 in a typical classroom, we can calculate the sound level as,

L = 10 log10(I/I0) = 10 log10(10^-7/10^-12) = 10 log10(10^5) = 50 dB

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It would take approximately 19.2 years for the asteroid to orbit once around the sun. But that none of the answer choices match the calculated value of approximately 19.2 years.

The period (T) of an orbit of a celestial body with semimajor axis (a) around the sun can be calculated using Kepler's third law:

T² = (4π² / GM) * a³

where G is the gravitational constant and M is the mass of the sun.

Plugging in the given value for the semimajor axis (a = 4 AU), we get:

T² = (4π² / (6.674 × 10⁻¹¹ m³/(kg s²) * 1.989 × 10³⁰ kg)) * (4 AU)³

T² = 3.652 × 10¹⁶ s²

Taking the square root of both sides, we get:

T = 6.04 × 10⁸ s

We can convert this time to years by dividing by the number of seconds in a year:

T = (6.04 × 10⁸ s) / (31,536,000 s/year)

T ≈ 19.2 years

Therefore, it would take approximately 19.2 years for the asteroid to orbit once around the sun. The closest answer choice is 16 years.

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The net work (W) done on the particle by the external forces during its motion can be expressed in terms of the initial (Ki) and final (Kf) kinetic energies as: [tex]W = ((1/2) \times m \times  vf^2) - ((1/2) \times  m \times  vi^2)[/tex]

To find the net work (W) done on the particle by the external forces during the particle's motion in terms of the initial (Ki) and final (Kf) kinetic energies, we will use the work-energy theorem. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

Step 1: Calculate the initial kinetic energy (Ki) and final kinetic energy (Kf).
Ki = (1/2) * m * vi²
Kf = (1/2) * m * vf²

Step 2: Calculate the change in kinetic energy (ΔK) as the difference between Kf and Ki.
ΔK = Kf - Ki

Step 3: According to the work-energy theorem, the net work (W) done on the particle by the external forces during its motion is equal to the change in kinetic energy (ΔK).
W = ΔK

Step 4: Substitute the expressions for Ki and Kf from step 1 into the equation for W from step 3.
W = ((1/2) * m * vf²) - ((1/2) * m * vi²)

In conclusion, the net work (W) done on the particle by the external forces during its motion can be expressed in terms of the initial (Ki) and final (Kf) kinetic energies as:  W = ((1/2) * m * vf²) - ((1/2) * m * vi²)

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Complete Question:

Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf. Work done on the particle by the external forces during the particle's motion. To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a . During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement is given by s=xf −xi.

The given statement " A series circuit is a current divider and a parallel circuit is a voltage divider circuit " is True

In a series circuit, the electric current is the same through each component, and the total current is equal to the sum of the currents through each component. Therefore, the current is divided among the components.

In a parallel circuit, the potential voltage across each component is the same, and the total voltage is equal to the sum of the voltages across each component. Therefore, the voltage is divided among the components.

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The force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is: 0.001 newtons per tesla

The force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is given by the formula F = (μI) / 2πr,

where μ is the permeability of free space, (4π x 10-7 N/A²)

I is current, and r is the radius of the loop.

In this case, the force is (4π x 10-7 x 5.5) / (2π x 0.1) = 0.001 N/T.

In other words, the force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is 0.001 newtons per tesla.

The formula for the force generated per unit field strength on a loop is derived from the fact that the force is a result of the magnetic field generated by the current flowing in the loop.

The magnitude of the magnetic field generated is proportional to the current and inversely proportional to the radius of the loop. Since the force is a product of the current and the magnetic field, it is proportional to the square of the current and inversely proportional to the square of the radius of the loop.

In summary, the force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is 0.001 newtons per tesla, given by the formula F = (μI) / 2πr, where μ is the permeability of free space (4π x 10-7 N/A²), I is current, and r is the radius of the loop.

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Torque is a measure of the twisting force that is produced when a force is applied to an object and is defined as the product of the force.

The distance from the pivot point to the point of application of the force, multiplied by the sine of the angle between the force vector and the vector from the pivot point to the point of application of the force.

In this case, the force of 90 N is applied perpendicular to the end of a wrench that is 0.5 m long. Assuming the force is applied at the end of the wrench, the distance from the pivot point to the point of application of the force is 0.5 m. Since the force is perpendicular to the wrench.

The angle between the force vector and the vector from the pivot point to the point of application of the force is 90 degrees. Using the formula for torque, the torque produced by the force is: Torque = force x distance x sin(angle)

Torque = 90 N x 0.5 m x sin(90)Torque = 45 Nm

Therefore, the torque produced by the force of magnitude 90 N that is exerted perpendicular to and at the end of a 0.5m long wrench is 45 Nm.

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According to the Stefan-Boltzmann law, the luminosity of a star is directly proportional to the fourth power of its temperature and its radius squared.

The formula for luminosity is:L = 4πR²σT⁴where L is the luminosity, R is the radius, T is the temperature, and σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²K⁴).To determine which stars would have the same luminosity as the sun, we need to compare their luminosity values using the given data. The radii and temperatures of five stars compared to the sun are as follows:Star A: R = 2R⊙, T = 6000 KStar B: R = R⊙, T = 3000 KStar C: R = 0.1R⊙, T = 6000 KStar D: R = 10R⊙, T = 3000 KStar E: R = 2R⊙, T = 15000 KSubstituting the values in the formula, we get:L⊙ = 4π(1²)(5.67 × 10⁻⁸)(5778⁴) ≈ 3.828 × 10²⁶ Wm¹²Star A: L = 4π(2²)(5.67 × 10⁻⁸)(6000⁴) ≈ 1.84 × 10³³ Wm¹²Star B: L = 4π(1²)(5.67 × 10⁻⁸)(3000⁴) ≈ 6.86 × 10²⁹ Wm¹²Star C: L = 4π(0.1²)(5.67 × 10⁻⁸)(6000⁴) ≈ 6.95 × 10²³ Wm¹²Star D: L = 4π(10²)(5.67 × 10⁻⁸)(3000⁴) ≈ 5.48 × 10³⁴ Wm¹²Star E: L = 4π(2²)(5.67 × 10⁻⁸)(15000⁴) ≈ 5.12 × 10³³ Wm¹²

The luminosity values of the stars are as follows:Star A: L ≈ 1.84 × 10³³ Wm¹²Star B: L ≈ 6.86 × 10²⁹ Wm¹²Star C: L ≈ 6.95 × 10²³ Wm¹²Star D: L ≈ 5.48 × 10³⁴ Wm¹²Star E: L ≈ 5.12 × 10³³ Wm¹²Comparing the luminosity values with that of the sun, we can see that stars A and E would have the same luminosity as the sun.

Therefore, the correct answer is: Stars A and E

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