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torisob [31]
3 years ago
6

If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di

fference between points A and B is
Physics
1 answer:
masya89 [10]3 years ago
8 0
The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5
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A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi
Norma-Jean [14]

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

8 0
3 years ago
A karate master strikes a board with an initial velocity of 10.0 m/s, decreasing to 1.0 m/s as his hand passes through the board
kenny6666 [7]

The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 1 m/s
  • Time (t) = 0.002 s
  • Mass (m) = 1 Kg
  • Force (F) = ?
<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

6 0
2 years ago
If a cotton ball is dropped from 12 meters with air resistance, what will be the velocity and acceleration at t = 1.00 s?
kifflom [539]
That depends on the weight, shape, size, density, and moisture content
of the cotton ball, as well as on the length, shape, thickness, and surface
texture of every little cotton fiber sticking out of it.

Now you know why we typically ignore air resistance when we work with
objects falling in gravity.
4 0
4 years ago
"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines a
sergey [27]

Answer:

B = 5.59x10⁹ T

Explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

F = qvBsin(\theta)

<u>We have:</u>

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!

7 0
3 years ago
PLEASE HELP ME!!!!!!!!
Ray Of Light [21]

Answer:

I believe C

Explanation:

i never learned this sorry

5 0
3 years ago
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