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zavuch27 [327]
3 years ago
13

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

ed from one electrode to the other, gains kinetic energy Ek. The distance between the electrodes is now changed to d. What is the gain in kinetic energy of an electron that is accelerated from one electrode to the other?
Physics
1 answer:
Alja [10]3 years ago
7 0

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

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Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210
kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

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Wavelength of light (\lambda )=565nm \approx 565\times 10^{-9}m

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

=\frac{2\lambda L}{b}

=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

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(b)Width of the first order bright fringe

Y_1=\frac{\lambda \times L}{b}

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What is the stretch when you pull with a force of 25 N on a spring with a spring constant of 8 N/m? *
Pani-rosa [81]

Hooke's Law

\tt F=k.\Delta x

k = spring constant

x = stretch

F = force

Input the value

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The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it i
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The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>
  • According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
  • By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = \omega  = \frac{2\pi N}{60}

\omega  = \frac{2\pi \times 1250}{60}

\omega  = 130.89 rad/s

Apparent weight is given by

W _a = M\omega^{2}R

W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225

W_a = 4.24084 N

Know more about Apparent weight brainly.com/question/14323035

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Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

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