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2 years ago

What is the frequency of the electromagnetic wave if the period is 1.54x10-15s

Physics

1 answer:

pogonyaev2 years ago

3 0

answer✿࿐

I was not able to write it here

so I did it somewhere else and attached the picture

i hope it helps

have a nice day

#Captainpower

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Pepsi [2]

Answer:

p.e=0.078kg×1/2×5.36m

p.e=0.913j

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3 years ago

Byron bought a keyless entry door lock that has the digits 0 through 9 on the keypad. He wants to choose a three-digit entry cod

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3 years ago

Read 2 more answers

The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel

aniked [119]

Answer:

= 2.52 x 10^ 6 m/s

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv) ÷ d

Here, d = distance between the two plates

q = charge of the charged particle

Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a = (q) (Δv) ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv) ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv) ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 = ut + 1/2 at^2 [proton}

put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron) ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

Note, e = 1.60 x 10^-19

m(electron) = 9.11 X 10^-31

m(proton) = 1.67 X 10^-27

Input these values into the formula above, initial speed, UI =

= 2.52 x 10^ 6 m/s

7 0

3 years ago

Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

MrMuchimi

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, $g=-9.81 m/s^2$ .