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3 years ago

An induced charge is when charged particles can be attracted to neutral objects. True or false?

Physics

1 answer:

Sindrei [870]3 years ago

5 0

Answer:

the answer is true Induction charging is a method used to charge an object without actually touching the object to any other charged object.

Explanation:

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A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo

kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0

3 years ago

Please help fill in the blank Earth’s axis passes through the (blank) and south poles.

LuckyWell [14K]

Answer:

North

Explanation:

Have a nice day :)

7 0

3 years ago

A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are mov

Kitty [74]

<span>net work = change in kinetic energy

for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction

10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²

now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2

since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)

10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8

they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>

5 0

3 years ago

I've got an energy and work problem. The premise of the problem is:

Alenkasestr [34]

Refer to the diagram shown below.

μ = the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)

6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)

7 0

3 years ago

how many revolutions are completed by a top spinning with a constant angular velocity of 3pi radians during a 20 sec time interv

patriot [66]

Answer:

Explanation:

Assuming the velocity is 3π radians <em>per second</em>, the top will spin through an angle of ...

(3π radians/s)(20 s) = 60π radians

Since each revolution is 2π radians, that is ...

(60π radians)/(2π radians/revolution) = 30 revolutions

7 0

3 years ago