A car moved 120km to the north. what is its displacement?

By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl

drek231 [11]

The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.

This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.

The more the distance between the charges, the less is the Electric Force.

The lesser the distance between the charges, the more is the Electric Force.

If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

Hence, the charge present determines a force to be attractive or repulsive.

Learn more about Coulomb Force here, brainly.com/question/15451944

#SPJ4

6 0

2 years ago

Which characteristic is common to all permanent magnets?

yanalaym [24]

I can't eliminate answers. Some of them are just wrong. A is incorrect. There is no such thing as a 1 pole magnet.

I wouldn't use B. If it is just a bar it is not a magnet.

C is the traditional answer

D is a space filler. It is just there to occupy a letter.

3 0

3 years ago

Read 2 more answers

Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of

WINSTONCH [101]

a₀). You know ...
       -- the object is dropped from 5 meters
   above the pavement;
         -- it falls for 0.83 second.

a₁). Without being told, you assume ...
   -- there is no air anyplace where the marshmallow travels,
   so it free-falls, with no air resistance;
   -- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .

b). You need to find how much LESS than 5 meters
   the marshmallow falls in 0.83 second.

c). You can use whatever equations you like.
   I'm going to use the equation for the distance an object falls in
   ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d). To see how this all goes together for the solution, keep reading:

The distance that an object falls in ' T ' seconds
when it's dropped from rest is

   (1/2 G) x (T²) .

On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall

   (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

   (5.00 - 3.38) = 1.62 meters

above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0

3 years ago

Calculate the displacement of a motorcycle that accelerates forward from rest to 7.0 m/s in 3.5 s.

Harrizon [31]

Answer:

12.25 meters

Explanation:

s=1/2(v+u)t

s= displacement

v= final velocity

u= initial velocity

t= time

7m/s+0m/s divide by 2= 3.5 m/s velocity Times 3.5 seconds= 12.25 meters

3 0

3 years ago

Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball i

alex41 [277]

The sum of the series allows to find the result for the total distance that the ball bounces is:

total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop height rebound

1 30 15

2 15 7.5

3 7.5 3.75

If we call the first term y₀

The first bounce can be found.

y₁ = $\frac{y_o}{2}$