The magnitude of the electric current is directly proportional to the _____________ of the electric field.

Answers

Answer 1
Answer:

Answer;

the potential difference

The magnitude of the electric current is directly proportional to the potential difference of the electric field

Explanation;

An electric current results from the collective movement of free charges under the effect of an electric field. An electric field exists and can be observed in the space around a single charge or a number of charges.

Electric fields cause charges to move. It stands to reason that an electric field applied to some material will cause currents to flow in that material. In other words, the current density is directly proportional to the electric field. The constant of proportionality σ is called the material’s conductivity.

Answer 2
Answer:

The Magnitude of the electric current is directly proportional to the voltage of the electric field.



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10. Which of the following statements is accurate?A. Sound waves passing through the air will do so as transverse waves, which vibrate vertically and still retain their horizontal positions. B. Compressions and rarefactions occur throughout a transverse wave. C. The wavelength of both traverse and longitudinal waves is measured parallel to the direction of the travel of the wave. D. Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.

Most of the stars in the Milky Way will end their lives as

Answers

 Most of the stars in the Milky Way will end their lives as white dwarfs
Most of the stars in the Milky Way will end their lives as white dwarfs.

Which of the outer planets is MOST like Earth? How?

Answers

Answer:

It is not impossible to answer the question, my apologies for “discrediting” the answer it would have to be Venus, the outer planet most similar to earth.

Explanation:

Venus is very similar to earth, as many studies, research, factual intellectual based concepts - the size, shape, surface gravity, average density, mass, average days of 24 hours, season changes similar to earth, and tilted axis. As well as similar climate make Venus the most accurate answer.

Hope this helps! Thank you for your time.

Answer: I would say that Saturn is most like Earth.  Not only does it have moons, like Earth, but the atmosphere is also denser like Earth. It also has windstorms, granite they are much more intense than Earth’s, but they do indeed happen. Saturn also has hydrogen gas; this gas is here on Earth supporting life in our ocean. Therefore, there is a chance that the conditions on Saturn could provide the essentials for there to be life.

in a series circuit: 1) the relationship between total resistance and individual resistances is 2) the relationship between total voltage and individual voltages is 3) the relationship between total current and individual currents is

Answers

1. The relationship between total resistance and individual resistances:

The total resistance(R_total) in a series circuit is equal to the sum of the individual resistances (R1, R2, R3, ...) connected in series. Mathematically, we can express this relationship as:

R_total = R1 + R2 + R3 + ...

In a series circuit, the total resistance increases as more resistors are added in series.

2. The relationship between total voltage and individual voltages:

In a series circuit, the total voltage (V_total) across the circuit is equal to the sum of the individual voltages (V1, V2, V3, ...) across each component. This relationship can be stated as:

V_total = V1 + V2 + V3 + ...

In a series circuit, the total voltage is equal to the sum of the individual voltage drops across each resistor.

3. The relationship between total current and individual currents:

In a series circuit, the total current (I_total) flowing through the circuit is the same at all points in the circuit.

This means that the current passing through each individual resistor is also equal to the total current. Mathematically, we can express this relationship as:

I_total = I1 = I2 = I3 = ...

In a series circuit, the current is constant throughout the circuit, and the same current flows through each component.

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Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is traveling in the same direction at 25 km/hr. Which of the following statements is true?A) The momentum before the collision is equal to the momentum after the collision.

B) The momentum before the collision is greater than the momentum after the collision.

C) The momentum before the collision is less than the momentum after the collision.

Answers

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car  A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as  m_(1) \ and\ m_(2)

Let the velocity of cars A and B are denoted as v_(1) \ and\ v_(2)

The momentum before collision is-

                                                  p_(i) =m_(1) v_(1) +m_(2) v_(2)

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of  linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are v'_(1) \ and\ v'_(2)

Hence the final momentum of the system is-

                                                        p_(f) = m_(1) v'_(1) +m_(2) v'_(2)

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e      

                              p_(i) =p_(f)

                               m_(1) v_(1) +m_(2)v_(2) =m_(1) v'_(1) +m_(2) v'_(2)

Hence the option A  is right.

A
This principle is known as the law of conservation of momentum.

Assignment: Ultrasound and Infrasound Research Exploration Ultrasound and infrasound are categories of sound with different frequencies. These categories of sound can be used for a variety of different applications. In this research assignment, you will take a look at the industrial applications of ultrasound and infrasound. Your essay should do the following things: • Define and describe in detail (and in your own words) ultrasound and infrasound • Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples. When conducting your research, remember to gather information from multiple sources. Start your research by defining ultrasound and infrasound. Then find ideas about industrial applications. The essay should be about 350-450 words, which is 1-1.5 typed pages, double-spaced, using 12-pt. font. Check the rubric to review how you will be graded for this assignment. plz no funny answers

Answers

Answer:

Infrasound vs. Ultrasound: Infrasound is sound that is below the lower limit of human hearing, below 20 Hz, and ultrasound is above the upper limit of human hearing, above 20,000 Hz.  Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes, graphing rock and oil developments underneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, a few creatures, for example, whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, water falls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

Explanation:

idk how many words this is but its a start for u to add on to and i hope this helps and its in my own words - pls mark me brainiest

Answer:

Ultrasound vs. Infrasound Research Exploration

Beyond the limit of human hearing, ultrasound is above 20,000 Hz. Under the limit of human hearing, infrasound is below 20 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes as well as graphing rock and oil developments beneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, creatures like whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, waterfalls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

(Not turned in yet, but this is what i have so far. Good luck 8th graders <33)

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The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 2006 the Earth took about 0.840 longer to complete 365 revolutions than it did in the year 1906.What was the average angular acceleration of the Earth during this time? Give your answer in rad/s2

Answers

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                Tₐ = 2π /ωₐ

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Final answer:

The average angular acceleration of the Earth from the year 1906 to 2006 would be -5.73 x 10^-20 rad/s^2. This value was obtained by finding the change in angular velocity and then dividing it by the elapsed time.

Explanation:

The question is asking for the average angular acceleration of the Earth from the year 1906 to 2006, during which the Earth's rotation rate decreased, causing the day to increase in duration by about 0.840 seconds.

To find the average angular acceleration, you first need to calculate the change in angular velocity, which can be found from the change in rotation time. One revolution (one day) is 2π radians, so the change in angular velocity is Δω = 2π/86400 s - 2π/(86400+0.840) s = -1.81 x 10^-10 rad/s.

The time interval from 1906 to 2006 is 100 years or about 3.16 x 10^9 seconds. Therefore, the average angular acceleration, α, which is the change in angular velocity divided by time, would be α = Δω/Δt = -1.81 x 10^-10 rad/s / 3.16 x 10^9 s = -5.73 x 10^-20 rad/s^2.

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