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2 years ago

A proton is at the origin. One electron is at the point (2m, 4m)

Physics

1 answer:

zmey [24]2 years ago

7 0

Answer:

Correct answer: F = 0 N

Explanation:

All three elementary charges are in the same straight line or direction.

Since the electron and proton are opposite charged, their force is attractive and directed toward the electron.

The forces are of the same intensity-magnitude and opposite directions and are canceled, so the resultant- net force by which the electrons act on the proton is zero.

God is with you!!!

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A 4 kg bird is flying with a velocity of 4 m/s. What is its kinetic energy?

Maslowich

32 kg m/s would be the kinetic energy.

3 0

2 years ago

A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a

Licemer1 [7]

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

v = w r

a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

v₁ = w 1,2 (boy)

v₂ = w 1.8 (girl)

whereby

v₂> v₁

reviewing the claims we have

a₁ = α 1,2

a₂ = α 1.8

a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

8 0

2 years ago

In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$