A uniform electric field is parallel to the x axis. In what direction can a charge be displaced in this field without any external work being done on the charge?

Answers

Answer 1

Answer:

Answer:

θ=π/2

Explanation:

The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2

Answer 2

Answer:

Final answer:

A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.

Explanation:

In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.

In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.

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How to Measure the mass of a coin using a triple beam balance

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Answer:

https://youtu.be/stW-C7F7QOg

After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

Answers

It will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

We can assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. We also assume that air resistance can be neglected.

Assuming that the package was dropped from rest at a height of h, the time it takes for the package to reach sea level can be calculated using the equation:

h = (1/2) * g * t²

where g is the acceleration due to gravity (9.8 m/s²) and t is the time it takes for the package to reach sea level.

Solving for t, we get:

t = sqrt(2h/g)

To convert the initial velocity of the package from km/hour to m/s, we can use the conversion factor:

1 km/hour = 0.2778 m/s

Therefore, the initial velocity of the package is:

v0 = 342 km/hour * 0.2778 m/s/km/hour = 95.0 m/s

if the package was dropped from a height of 5000 meters, the time it takes for the package to reach sea level is:

h = 5000 m

t = sqrt(2h/g) = sqrt(2*5000/9.8) = 32.0 seconds

Therefore, it will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

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Final answer:

The time a dropped package takes to reach sea level from a plane is determined by its vertical motion. If the package retains only horizontal velocity when released, the time taken would be calculated using the height from which the object is dropped. However, to give a numerical value of time, we need to know the exact height.

Explanation:

The time it takes for the package dropped from the plane to reach sea level is determined exclusively by the package's vertical motion, assuming the package does not face air resistance. Specifically, the time of flight for a projectile launched and landing at the same elevation is governed by the equation: t = 2*v/g, where v represents the initial vertical velocity and g is the acceleration due to gravity. From the scenario, it seems the package retains only horizontal velocity when released since it's dropped down directly rather than being thrown downward, hence rendering initial vertical velocity as zero. Simply put, the package only begins to accelerate in the vertical direction once it's dropped, meaning the time taken would be calculated using the equation: t = √(2h/g), h being the height from which the object is dropped.

In the provided context, unfortunately, we need the height from which the package is dropped to give a specific numerical value of the time in seconds. If we knew the height of the plane at the time the package was dropped, we'd recalculate the time in seconds more precisely.

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Experts in model airplanes develop a supersonic plane to scale, it moves horizontally in the air while it is conducting a flight test. The development team defines that the space that the airplane travels as a function of time is given by the function: e (t) = 9t 2 - 6t + 3 Determine what acceleration the scale airplane has (Second derivative).

Answers

Explanation:

e(t) = 9t² − 6t + 3

The velocity is the first derivative:

e'(t) = 18t − 6

The acceleration is the second derivative:

e"(t) = 18

PLEASE HELP IT'S DUE IN LIKE 2 MINUTES

Answers

Answer:

1kg

Explanation:

this box is the smallest and weighs the least. Hope this helps :]

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

Answers

Here is the full question

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi ((D)/(2))^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= (A)/(N) =(\pi ((D)/(2))^2 )/(10, 000)$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= (d)/(\pi)$

$r_e = \sqrt{(d)/(\pi) }$

replacing d = $(\pi ((D)/(2))^2 )/(10, 000)$ in the equation above; we have:

$r_e = \sqrt{((\pi ((D)/(2))^2 )/(10, 000))/(\pi) }$

$r_e = \sqrt{(((D)/(2))^2 )/(10, 000)}$

$r_e = \sqrt{(((100,000)/(2))^2 )/(10, 000)}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.