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11Alexandr11 [23.1K]
4 years ago
8

Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity.

Physics
1 answer:
Sholpan [36]4 years ago
7 0
The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
Let F be the viscous force due to air molecules, B be buoyant force due to air and W be the weight of falling object. Initially, the velocity of falling object and hence the viscous force F is zero and the object is accelerated due to force
(W-B). Because of the acceleration the velocity increases and accordingly the viscous force also increases. At a certain instant, the viscous force becomes equal to W-B. The net force then becomes zero and the object falls with constant velocity. This constant velocity is called terminal velocity.
Thus at terminal velocity, air resistance and force of gravity becomes equal.
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"Which of the following is true about the atom shown? Choose all that apply.
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3 years ago
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The speed of a sound wave traveling through air is 400 meters per second. The wavelength of the wave is 25 meters What is the
sweet [91]

Answer:

16Hz

Explanation:

Given parameters:

Speed of sound  = 400m/s

Wavelength  = 25m

Unknown:

Frequency of the wave  = ?

Solution:

To solve this problem;

      V  = F ∧  

V is the velocity

F is the frequency

∧ is the wavelength

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        F  = 16Hz

7 0
3 years ago
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Which of Newton's laws accounts for the following statement? "A force cannot act alone."
juin [17]
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8 0
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A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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