1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MrRissso [65]
3 years ago
11

Why are objects that fall near Earth's surface rarely in free fall?

Physics
1 answer:
sveta [45]3 years ago
6 0

Answer: Air exerts forces on falling objects near Earth's surface.

Explanation: Objects falling near the earth surface are rarely in free fall due to the force exerted by air on falling object near the earth( air resistance).

The acceleration of free fall tend to pull the body towards the earth surface while air resistance (drag) tends to act in the opposite direction.

Given that the weight of the body is always  constant.

The drag acts in the upward direction, thereby negating the downward weight of the object.

From Newton's Second law:

Force = mass × acceleration due to gravity

Acceleration = force / mass

The net force acting on the object becomes :

(Downward weight - upward drag)

This hampers the free fall of the object due to gravity.

You might be interested in
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
You go rock climbing with a pack that weighs 70 N and you reach a height of 30 m. how much work did you do to lift your pack? If
Vikentia [17]
When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).
In order to climb, you need to work against work done by gravity on the pack.
Hence work done by you = work done by gravity on pack 
                                        = Force x displacement = 70 x 30 = 2100 J.
 
So you need to do 2100 joules of work to lift your pack.

Power is the rate of work done.
Therefore power = work done by you/time(in seconds)
                            =     2100/600 =3.5 watts 
8 0
3 years ago
Why does the total amount of energy before and after any energy transformations remain the same?
Georgia [21]
The amount of energy before and after any energy transformations remain the same because energy cannot be created or destroyed. From the law conservation of energy; any time energy is transferred between two objects, or converted from one form into another, no energy is created and none is destroyed. The total amount of energy involved in the process remains the same.
3 0
3 years ago
Read 2 more answers
Gerry is looking at salt under a powerful microscope and notice a crystalline structure.what can be known about the salt sample
Natali [406]

Well, if the salt that Gerry's looking at under a powerful microscope has a crystalline structure, then that's saying that salt is technically a solid.

(I hope that this is an answer you were looking for)

6 0
3 years ago
What is the term used to label the energy levels of electrons?
mel-nik [20]
Electron volts...........
4 0
3 years ago
Other questions:
  • Please help!!!! due today!!
    6·1 answer
  • a trombone can be modeled like an open closed air tube. the trombone plays a fifth harmonic of 159 hz. the speed of sound is 343
    6·1 answer
  • It takes brooke 10minutes to run 1 mile what is her speed in miles per minutes
    7·1 answer
  • Ed records the temperature of two pots of water. Pot X
    12·2 answers
  • Where is 3cm on a ruler
    11·1 answer
  • Prior to the music CD, stereo systems had a phonographic turntable on which vinyl disk recordings were played. A particular phon
    15·1 answer
  • A positive charge of 0.026 C moves horizontally to the right at a speed of 443.592 m/s and enters a magnetic field directed vert
    6·1 answer
  • Discuss the relationship between electric and magnetic fields
    14·1 answer
  • Dimension of radius of sphere​
    7·1 answer
  • What is the λ if v = 75 m/s and ƒ = 25 Hz?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!