All categories

3 years ago

Why are objects that fall near earth’s surface rarely in free fall?

Physics

1 answer:

Sloan [31]3 years ago

3 0

Answer:

Because of the presence of air resistance

Explanation:

When an object is in free fall, ideally there is only one force acting on it:

- The force of gravity, W = mg, that pushes the object downward (m= mass of the object, g = acceleration of gravity)

However, this is true only in absence of air (so, in a vacuum). When air is present, it exerts a frictional force on the object (called air resistance) with upward direction (opposite to the motion of free fall) and whose magnitude is proportional to the speed of the object.

Therefore, it turns out that as the object falls, its speed increases, and therefore the air resistance acting against it increases too; as a result, the at some point the air resistance becomes equal (in magnitude) to the force of gravity: when this happens, the net acceleration of the object becomes zero, and so the speed of the object does not increase anymore. This speed reached by the object is called terminal velocity.

You might be interested in

What is true about high-level nuclear waste?

fenix001 [56]

the answer is b. it's commercially reprocessed

3 0

4 years ago

Read 2 more answers

A gas-turbine power plant operating on the simple Brayton cycle has a pressure ratio of 7. Air enters the compressor at 0°C and

Ivahew [28]

Answer:

Answers of the Both parts are in the following attachment

Explanation:

4 0

3 years ago

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

$\frac{d'_1}{d'_2}=\frac{f_1}{f_2}$

Magnification of first lens

$M_2=-\frac{d'_1}{d_1}$

and

$M_2=\frac{h'_1}{h_1}$

Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

$\frac{d'_1}{d_2'}=\frac{f_1}{f_2}$

So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

∴ Focal length should this camera's lens is 24.71 mm

6 0

3 years ago

Two positive charges are labeled q Subscript 1 baseline and q Subscript 2 baseline are 0.1 m apart. Two positively charged parti

Snowcat [4.5K]

Answer:

Explanation:

Let that point be at a distance x from q1

Then Kq1/x^2= Kq2/ (s-x)^2

Taking square roots and simplifying, x =s /[1+(q2/q1)^0.5]

Assuming an identical distance, the rigidity of Q on 2Q is equivalent in value to the rigidity of 2Q on Q. for that reason, had the area R been stored an identical, the two forces could be equivalent. inspite of the shown fact that, via fact the area is being decreased, we could constantly consult with the equation we use to calculate those forces: F = ok(Q1xQ2)/(R^2) because R is squared and is being halved, the final result's that's it being divided by potential of a million/4. for that reason, the rigidity would be expanded by potential of four, and be 4F.

4 0

4 years ago

In the following reaction, 24 g of CH4 react with 96 g of Oz to form 66 g of CO2

Brut [27]

Answer:

54 g

Explanation:

Mass is conserved. Mass of reactants = mass of products.

24 g + 96 g = 66 g + x

x = 54 g

8 0

3 years ago