5 minerals used to make phones

An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only

topjm [15]

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² = 1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities are inversely proportional to square root of their masses .

4 0

3 years ago

What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25

Rom4ik [11]

Acceleration of the ball is $50 m/s^2$

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

$a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

2 years ago

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge

Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

= k λ dx .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫ k λ .15 / (x² + .15² )³/² dx

= k λ x L / .15 √( L² / 4 + .15² )

6 0

3 years ago

An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what

Katarina [22]

Answer:

Hiii how are you doing??I don't understand that

3 0

2 years ago

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your an

AveGali [126]

Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

$\phi=123.75$