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Ymorist [56]
3 years ago
14

Donna stretched a spring to increase its length. Nearby in an electrical device, a positive charge attracted a negative charge.

Which statement is true about the spring and the negative charge?
A. They both experienced contact forces.
B. They both experienced non-contact forces.
C. The spring experienced a contact force and the charge experienced a non-contact force.
D. The spring experienced a non-contact force and the charge experienced a contact force.
Physics
1 answer:
timurjin [86]3 years ago
7 0

C. The spring experienced a contact force and the charge experienced a non-contact force

Explanation:

In physics, there are two types of forces:

  • Contact forces: contact forces are those where a contact between the objects involved is needed, in order to have the force. Examples of contact forces are the force of friction (the contact between the two surfaces is needed), or the elastic force in a spring (in fact, the spring needs to be physically in contact with something that stretches it)
  • Non-contact forces: non-contact forces are those where no contact is needed between the objects, so they can be exerted at a distance. Examples are the gravitational force (two masses exert a force on each other even from a distance) or the electric force (two charges exert a force on each other even from a distance).

Therefore, in this example:

  • The spring experiences a contact force
  • The two charges experience a non-contact force

So the correct answer is

C. The spring experienced a contact force and the charge experienced a non-contact force

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

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