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Arada [10]
3 years ago
13

What happens to the speed of an object that is dropped in free fall

Physics
1 answer:
bagirrra123 [75]3 years ago
8 0
<span>
If it's not falling through air, water, smoke, or anything else,
and gravity is the only force on it, then its speed increases
at a constant rate ... 9.8 meters per second for every second
it falls.  (That's the number on Earth.  It's different in other places.) </span>
You might be interested in
Help with either I don’t understand this I know W=f*d but don’t get how it applies here
rosijanka [135]

Answer:

5) Displacement = +3.125 m

Displacement is in the same direction as the force vector.

6) Force = -53.89 N

Force is in an opposite direction relative to the displacement.

Explanation:

5) We are given;

Force; F = 160 N.

Workdone; W = +500 J

Now, formula for workdone is;

W = Force × displacement

Thus, displacement = Work/force

Displacement = 500/160

Displacement = +3.125 m

Thus, displacement is in the same direction as the force vector.

6) We are given;

Displacement; d = 18 m.

Workdone; W = -970 J

Like in the first answer above,

Workdone = Force × Displacement

Thus;

Force = Workdone/Displacement

Force = -970/18

Force = -53.89 N

Since force is negative and displacement is positive, it means force is in an opposite direction relative to the displacement.

3 0
3 years ago
A roller coaster track is 3000 meters long. It takes 100 seconds to travel once around the roller coaster. What is the average s
Svetllana [295]
All you have to do is divide 3000 by 100, Its 30
6 0
3 years ago
Read 2 more answers
1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7
xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

11.5x31540000=362710000

7 0
2 years ago
Which of the following is best described as an inclined plane twisted into a spiral?
Margarita [4]

the answer is A or B

5 0
3 years ago
A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the
Aleksandr-060686 [28]

Answer:

(a):  The magnitude of the electric force on the small sphere = \dfrac{q\sigma}{2\epsilon_o}.

(b): Shown below.

Explanation:

<u>Given:</u>

  • m = mass of the small sphere.
  • q = charge on the small sphere.
  • L = length of the silk fiber.
  • \sigma = surface charge density of the large vertical insulating sheet.

<h2>(a):</h2>

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

\rm E = \dfrac{\sigma}{2\epsilon_o}.

<em>where,</em>

\epsilon_o is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

\rm F_e=qE.

Thus, the magnitude of the electric force on the small sphere is given by

\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

<h2>(b):</h2>

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

\rm \tan \theta = \dfrac{F_e}{W}.

<em>where,</em>

  • \rm F_e = electric force on the sphere, acting along left.
  • \rm W = weight of the sphere, acting vertically downwards.

<em />

\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .

g is the acceleration due to gravity.

6 0
3 years ago
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