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MAVERICK [17]
3 years ago
5

A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will

be
​
Physics
1 answer:
Pavel [41]3 years ago
8 0

Answer:

Final Velocity = √(eV/m)

Explanation:

The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference

W = (2e) × V = 2eV

And this work is equal to change in kinetic energy

W = Δ(kinetic energy) = ΔK.E

But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0

ΔK.E = ½ × (mass) × (final velocity)²

(Velocity)² = (2×ΔK.E)/(mass)

Velocity = √[(2×ΔK.E)/(mass)]

ΔK.E = W = 2eV

mass = 4m

Final Velocity = √[(2×W)/(4m)]

Final Velocity = √[(2×2eV)/4m]

Final Velocity = √(4eV/4m)

Final Velocity = √(eV/m)

Hope this Helps!!!

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A body of mass 2 kg at O has an initial velocity of 3m/s along OE and it is subjected to a force of 4N perpendicular to OE the d
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Perpendicular acceleration:
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Perpendicular distance:

s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
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Horizontal distance:
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4 years ago
Please help!!! A river has a constant current of 3 km per hour. If a motorboat, capable of maintaining a constant speed of 20km
PilotLPTM [1.2K]
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have: 

<span>theta = arcsin(3/20) = approx. 8.63° </span>

<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>

<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>

<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
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4 years ago
When a wave is bent by traveling from one medium to another
Natalka [10]

Answer:

B. Refraction

Explanation:

Jope this helps

6 0
3 years ago
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31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
posledela

Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

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b.

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a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

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