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Likurg_2 [28]
3 years ago
8

What two things will affect size of the electric force between two objects?​

Physics
1 answer:
tatyana61 [14]3 years ago
4 0

<u>Answer:</u>

The two things that affect the size of the electric force between two objects are "amount of charge on each object and distance between the charges".

<u>Explanation:</u>

The magnitude and sign of the electric force are estimated by the electric charge in "Coulombs Law" inspite of mass an object which is favorable in gravitational force. Therefore charge is the property of electromagnetism which influences the charged objects' motion.

More the charge on object more will be the electric force, as bigger charge cloud have more force than smaller.The electric force is inversely proportional to the square of the distance between the two charges. This showcase weaker attraction or repulsion as the distance increases between the two charges.

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What is the velocity of an object that has been in free fall for 1.5s?
Crank

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

8 0
3 years ago
A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal
Lynna [10]

Wow !  This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

                          (speed-squared)  /  (radius of the circle)  .

Notice that we won't need to use the mass of the train.

We know the radius of the track.  We don't know the trains speed yet,
but we do have enough information to figure it out.  That's what we
need to do first.

Speed  =  (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) =  2.4π  meters

10 laps =  24π  meters.

Time = 1 minute 20 seconds  =  80 seconds

The trains speed is  (distance) / (time)

                               =  (24π meters) / (80 seconds)

                               =        0.3 π  meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

                                 <span> (speed)²  /  (radius)

                           =    (0.3π m/s)²  /  (1.2 meters)

                           =    (0.09π m²/s²)  /  (1.2 meters)

                           =    (0.09π  /  1.2)   m/s²

                           =          0.236  m/s²  .        (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well,       Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force.       </span>
4 0
2 years ago
Why does a box on the seat of a car slide around on the c one the car speeds up slows down or turns a corner?
Feliz [49]
Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.
8 0
2 years ago
The image below shows four points on the orbit of a comet around the Sun.
Firlakuza [10]

Answer:

<u>Point </u><u>A</u>

Explanation:

And the strongest force of gravity is point C

3 0
2 years ago
by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th
xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}

\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}

\displaystyle F'=9k\frac{q_1q_2}{d^2}

\displaystyle F'=9F

The force is now 9 times the original force

8 0
2 years ago
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