when the distance between two charges is halved, the electrical force between the charges is reduced by 1/4. quadruples. halves. doubles. none of the above choices are correct.

Answers

Answer 1

When the distance between two charges is halved, the electrical force between the charges quadruples. This is due to the inverse square relationship between distance and electrical force, which means that when distance is halved, the force increases by a factor of 4.



The electrical force between the charges quadruples when the distance between them is halved. This is due to Coulomb's Law, which states that the electrical force (F) between two charges (q1 and q2) is directly proportional to the product of the charges and inversely proportional to the square of the distance (r) between them. Mathematically, it can be expressed as:

F = k * (q1 * q2) / r^2

When the distance (r) is halved, the denominator (r^2) becomes 1/4 of its original value, which causes the electrical force (F) to be 4 times greater, or quadruple.

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Related Questions

maxwell's equations are a complete description of electric and magnetic fields. how many equations are there?

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Maxwell's equations are a complete description of electric and magnetic fields. There are four equations in Maxwell's equations. These four equations are:

1. Gauss's Law for Electric Fields: Describes the relationship between electric charges and the electric field produced by them.
2. Gauss's Law for Magnetic Fields: States that there are no magnetic monopoles, and the magnetic field lines are always closed loops.
3. Faraday's Law of Electromagnetic Induction: Describes the induced electromotive force (EMF) in a closed circuit produced by a changing magnetic field.
4. Ampere's Law with Maxwell's Addition: Relates the magnetic field around a closed loop to the electric current passing through the loop and the rate of change of the electric field.

These four equations collectively provide a comprehensive description of electric and magnetic fields and their interactions.

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A compound microscope is a two-lens system used to look at very small objects. Which of the following statements is correct? The objective lens is a short focal length, convex lens and the eyepiece functions as a simple magnifier. The objective lens is a long focal length, convex lens and the eyepiece functions as a simple magnifier. The objective lens and the eyepiece both have the same focal length and both serve as simple magnifiers. The objective lens is a short focal length, concave lens and the eyepiece functions as a simple magnifier. The objective lens is a long focal length, concave lens and the eyepiece functions as a simple magnifier.

Answers

The objective lens is a long focal length, convex lens and the eyepiece functions as a simple magnifier is the correct statement about a compound microscope. Option b is correct.

In a compound microscope, the objective lens is a long focal length, convex lens that produces an inverted, magnified real image of the specimen. The eyepiece, on the other hand, functions as a simple magnifier, which further magnifies the real image produced by the objective lens and forms a virtual image that can be viewed by the observer's eye. Therefore, option b is the correct statement.

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--The complete question is, A compound microscope is a two-lens system used to look at very small objects. Which of the following statements is correct?

a. The objective lens is a short focal length, convex lens and the eyepiece functions as a simple magnifier.

b. The objective lens is a long focal length, convex lens and the eyepiece functions as a simple magnifier.

c.  The objective lens and the eyepiece both have the same focal length and both serve as simple magnifiers.--

T/F : Staleness and burnout are not associated with overtraining.

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False. Staleness and burnout are often associated with overtraining, which occurs when an individual exceeds their capacity to recover from intense physical training or activity.

Overtraining can lead to physical and psychological symptoms, including decreased performance, fatigue, irritability, and decreased motivation. It is important for individuals to listen to their bodies and take rest and recovery periods to prevent overtraining and associated symptoms.

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voyager 1 is a space probe launched by nasa in 1977 and is the farthest human-made object. it experiences negligible gravity. voyager 1 is propelled by thrusters but will run out of fuel by 2040. what will happen to voyager 1 after this date?multiple select question.the velocity of voyager 1 will remain unchanged.voyager 1 will slow down from the velocity it will have when the fuel runs out.voyager 1 will immediately stop.voyager 1 will continue moving with the speed it will have when the fuel runs out.

Answers

Voyager 1 will continue moving with the speed it will have when the fuel runs out. The probe is traveling through the vacuum of space, where there is negligible gravity and no significant air resistance to slow it down.

Without the ability to adjust its trajectory, Voyager 1 will continue on its current path indefinitely unless it encounters a gravitational field that alters its trajectory. The probe may eventually drift off course and potentially collide with other celestial objects in its path. While Voyager 1 will continue to communicate data to Earth until its systems eventually fail, it will eventually become just another piece of space debris, floating silently through the cosmos.

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How might you utilize your recently acquired knowledge from PSIO 305 to make a daytime hike in the 49 degree Celsius Mojave Desert non-deadly?
Question options:
all of the above
decrease surface area to maximize convection
take medication to suppress aldosterone
drink lots of water to increase evaporative water loss
take off your shirt to increase radiative heat loss

Answers

One should drink a lot of water to maximise evaporative water loss in order to make a day walk in the Mojave Desert, where the temperature is 49 degrees Celsius, not fatal.

This will support hydration levels maintenance and temperature control. Wearing loose, light-colored clothing, taking breaks in the shade, and taking off your shirt to promote radiative heat loss can also help reduce surface area to maximise convection. However, it is not advised to take aldosterone-suppressing medication without a doctor's supervision.

Human physiology, which is covered in PSIO 305, teaches students how the body functions in various situations. The body may be subjected to intense heat in the Mojave Desert, which can cause dehydration and disorders associated with heat. Staying hydrated and controlling body temperature through sweating and evaporative water loss are crucial to avoiding this. Reduce heat absorption by dressing appropriately, taking rests in the shade, and using air conditioning. Aldosterone is a hormone that controls electrolyte balance; nevertheless, taking medicine to inhibit it might have consequences and is not advised without a doctor's supervision.

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One should drink a lot of water to maximise evaporative water loss in order to make a day walk in the Mojave Desert, where the temperature is 49 degrees Celsius, not fatal.  Option d.

This will support hydration levels maintenance and temperature control. Wearing loose, light-colored clothing, taking breaks in the shade, and taking off your shirt to promote radiative heat loss can also help reduce surface area to maximise convection. However, it is not advised to take aldosterone-suppressing medication without a doctor's supervision.

Human physiology, which is covered in PSIO 305, teaches students how the body functions in various situations. The body may be subjected to intense heat in the Mojave Desert, which can cause dehydration and disorders associated with heat. Staying hydrated and controlling body temperature through sweating and evaporative water loss are crucial to avoiding this.

Reduce heat absorption by dressing appropriately, taking rests in the shade, and using air conditioning. Aldosterone is a hormone that controls electrolyte balance; nevertheless, taking medicine to inhibit it might have consequences and is not advised without a doctor's supervision.

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Full Question: How might you utilize your recently acquired knowledge from PSIO 305 to make a daytime hike in the 49 degree Celsius Mojave Desert non-deadly?

a. decrease surface area to maximize convection

b. all of the above

c. take medication to suppress aldosterone

d. drink lots of water to increase evaporative water loss

e. take off your shift to increase radiative heat loss

A proton accelerates from rest in a uniform electric field of 691 N/C. At some time later, it’s speed is 2. 30 x 10^6 m/s. (a) What is the magnitude of its acceleration? (b) How long does it take the proton to reach this speed

(c) How far has it moved in this time interval?

(d) What is its kinetic energy at the later time?

Mass of proton: 1. 6726x10^-27

Fundamental charge:

1. 602 x10^-19

Answers

The proton experiences an acceleration of [tex]$6.60\times10^{10} \text{m/s}^2$[/tex] in a uniform electric field of 691 N/C, and it takes [tex]$3.48\times10^{-5}$[/tex] s to reach a velocity of [tex]$2.30\times10^{6}$[/tex] m/s. During this time, the proton travels a distance of [tex]$4.36\times10^{-10}$[/tex] m and has a kinetic energy of [tex]$3.07\times10^{-12}$[/tex] J.

(a) The magnitude of the acceleration experienced by the proton can be determined by using the equation for the force on a charged particle in an electric field, which is F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength. For a proton, the charge is equal to the fundamental charge, which is [tex]$1.602\times10^{-19} \text{C}$[/tex]. Therefore, the force on the proton is [tex]$F = (1.602\times10^{-19} \text{C})(691 \text{N/C}) = 1.106\times10^{-16} \text{N}$[/tex]

The acceleration of the proton can be determined using the equation F = ma, where m is the mass of the proton. Thus, [tex]$a = F/m = \dfrac{1.106\times10^{-16} \text{N}}{1.6726\times10^{-27} \text{kg}} = 6.60\times10^{10} \text{m/s}^2$[/tex].

(b) To find the time it takes for the proton to reach the given speed, we can use the kinematic equation v = u + at, where u is the initial velocity (which is 0 m/s), v is the final velocity ([tex]$2.30\times10^{6} \text{m/s}$[/tex]), a is the acceleration ([tex]$6.60\times10^{10} \text{m/s}^2$[/tex]), and t is the time. Rearranging this equation gives [tex]$t = \dfrac{v-u}{a} = \dfrac{2.30\times10^{6} \text{m/s}}{6.60\times10^{10} \text{m/s}^2} = 3.48\times10^{-5} \text{s}$[/tex].

(c) The distance the proton has moved in this time interval can be calculated using the kinematic equation [tex]$s = ut + \dfrac{1}{2}at^2$[/tex], where s is the distance traveled. Substituting the known values, we get [tex]$s = \dfrac{1}{2}(6.60\times10^{10} \text{m/s}^2)(3.48\times10^{-5} \text{s})^2 = 4.36\times10^{-10} \text{m}$[/tex]

(d) The kinetic energy of the proton can be calculated using the equation [tex]$KE = \dfrac{1}{2}mv^2$[/tex], where KE is the kinetic energy, m is the mass of the proton, and v is the velocity of the proton. Substituting the known values, we get [tex]$KE = \dfrac{1}{2}(1.6726\times10^{-27} \text{kg})(2.30\times10^{6} \text{m/s})^2 = 3.07\times10^{-12} \text{J}$[/tex].

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another capacitor, identical to the original, is added in series to the circuit described in the passage. compared to the original circuit, the equivalent capacitance of the new circuit is:

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The equivalent capacitance of the new circuit with an identical capacitor added in series is half of the original circuit's capacitance.

When a second capacitor, identical to the original, is added in series to the circuit, the equivalent capacitance of the new circuit is reduced. This is because the total capacitance in a series circuit is always less than the individual capacitances. The formula for calculating the equivalent capacitance of a series circuit is:

[tex]1/Ceq = 1/C1 + 1/C2 + ... + 1/Cn[/tex]

Where C1, C2, ..., Cn are the capacitances of the individual capacitors.

Adding another capacitor in series to the circuit means that the equivalent capacitance will be smaller, and the total charge stored in the circuit will be less. This will affect the behavior of the circuit when connected to a voltage source, as it will take less time to charge and discharge.

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A 60-kg swimmer suddenly dives horizontally from a 150-kg raft with a speed of 1. 5 m/s. The raft is initially at rest. What is the speed of the raft immediately after the diver jumps if the water has negligible effect on the raft?

Answers

The speed of the raft immediately after the diver jumps is 0.6 m/s.

After the swimmer jumps, the momentum of the system is still conserved, but it is no longer zero, since the swimmer is now moving. We can use the equation:

(m1v1 + m2v2)before = (m1v1 + m2v2)after

We want to solve for v2, velocity of the raft immediately after the jump.

Before jump, velocity of  raft is zero, so we can simplify  equation to:

m1v1 = m2v2

Substituting in  values we know, we get:

60 kg * 1.5 m/s = 150 kg * v2

Simplifying, we get:

v2 = (60 kg * 1.5 m/s) / 150 kg = 0.6 m/s

So the speed of the raft immediately after the diver jumps is 0.6 m/s.

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