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harina [27]
3 years ago
8

Do you think the electron will feel a force towards the proton, or away from it due to the proton's electric field? Given that,

how do you expect the electron and proton to move?
Physics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

Yes, electron will move towards proton due to the proton's electric field. They moved towards each other.

Explanation:

Due to proton's field electron experienced force toward proton.

Since electron is negative charge it  experience force toward proton due to positive charge of proton  and proton experience force toward electron , therefore both moves towards each other . Movement of both electron and proton is attractive.

Note:- unlike charges always attracts Each other and like charges moves away from each other.

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Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.
o-na [289]

Answer:

Current = 10 Amperes.

Explanation:

Given the following dat;

Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C

Time = 1 hour to seconds = 60*60 = 3600 seconds

To find the current;

Quantity of charge = current * time

Substituting in the equation

36000 = current * 3600

Current = 36000/3600

Current = 10 Amperes.

6 0
2 years ago
Explain the difference between cool air and warm air
Naddik [55]

Answer: ok

Explanation:

The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less.

3 0
2 years ago
Suppose of methane are burned in air at a pressure of exactly and a temperature of . Calculate the volume of carbon dioxide gas
Licemer1 [7]

There are supposed to be numbers in these places:

" _____ of methane"

" ... pressure of exactly _____ "

" ... temperature of _____ "

I've seen a lot of questions with missing numbers before, but I think this is the first time I ever saw a question where even the number of significant digits to round the answer to is missing.

"Round your answer to _____ significant digits."

4 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo
Vlad1618 [11]

Answer:

Tension T1 is less than tension T2.

T1 < T2

Explanation:

According to given data,

mass of box A ( mA) is grater than mass of box B (mB)

we can write,

m(A) > m(B)

Newton's second law states that:

Tension of object is directly proportional to the mass of the system.

T ∝ m

here Boxes A and B are being pulled to the right on a frictionless surface,

so Tension T1 generates due to the mass of box A m(A)

and Tension T2 arises due to mass of the system m(A) + m(B)

Thus tension T1 will be less than tension T2

T1 < T2

learn more about Tension force here:

<u>brainly.com/question/13175014</u>

<u />

#SPJ4

8 0
2 years ago
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