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4 years ago

Air resistance is an example of what type of friction?

1 answer:

4 0

Well, Air resistance is a special type of friction (you cannot classify it in other categories). That force of air-resistance is often observed to oppose the motion of the object,( like every other frictional forces)

Hope this helps!

You might be interested in

Which of the following masses is the smallest?

yarga [219]

The smallest unit of matter is a atom

4 0

3 years ago

Read 2 more answers

How long would it take for a person to sprint from the 0m line to the 100m line if they are traveling forward at 6m/s?

valina [46]

Answer:

<em><u>This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.</u></em>

Explanation:

<h3>Derive the kinematic equations for constant acceleration using integral calculus.</h3><h3>Use the integral formulation of the kinematic equations in analyzing motion.</h3><h3>Find the functional form of velocity versus time given the acceleration function.</h3><h3>Find the functional form of position versus time given the velocity function.</h3>

8 0

3 years ago

.. A 15.0-kg fish swimming at 1.10 m>s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effec

stira [4]

Answer:

(a) 0.846 m/s

(b) 2.097J

Explanation:

Parameters given:

Mass of big fish, M = 15 kg

Mass of small fish, m = 4.5 kg

Initial speed of big fish, U = 1.1 m/s

Initial speed of small fish, u = 0 m/s (it is stationary)

(a) We apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

Since both fish have the same final speed, V, (the small fish is in the mouth of the big fish), we have:

MU + mu = (M + m)*V

(15 * 1.1) + (4.5 * 0) = ( 15 + 4.5) * V

16.5 = 19.5V

=> V = 16.5/19.5

V = 0.846 m/s

The speed of the large fish after the meal is 0.846 m/s.

(b) We need to find the change in Kinetic energy of the entire system to find the total mechanical energy dissipated.

Initial Kinetic energy:

KEini = (½ * M * U²) + (½ * m * u²)

KEini = (½ * 15 * 1.1²) + (½ * 4.5 * 0²)

KEini = 9.075 J

Final Kinetic Energy:

KEfin = (½ * M * V²) + (½ * m * V²)

KEfin = (½ * 15 * 0.846²) + (½ * 4.5 * 0.846²)

KEfin = 5.368 + 1.610 = 6.978 J

Change in kinetic energy will be:

KEfin - KEini = 9.075 - 6.978

ΔKE = 2.097 J

The energy dissipated in eating the meal is 2.097 J

5 0

3 years ago

Bob is watching Anna fly by in her new high-speed plane, which Anna knows to be 60 m in length. As a greeting, Anna turns on two

vlada-n [284]

Answer:

0.196*c

Explanation:

A good thing to remember is that once you have the position and the time in one frame we can figure out corresponding position and time in another frame in other frames given their relative velocities.

Anna moves in (S') frame and we are given two events synchronized in her frame with the distance between them, Hence using Lorentz transformation to show which event Bob frame (S) will see first:

Hence,

$dt = y (dt' + \frac{v*dx}{c^2} ) \\\\dt' = 0\\dx' = 60 m\\\\t_2 - t_1 = \sqrt{\frac{1}{(1 - v^2/c^2) } } * (\frac{v*(x_2 - x_1)}{c^2} )$

we are assuming that dx' > 0 , hence we are saying that light emitted from front end of the plane is event number two and that coming from the back end is event number one. As the time in Bob's frame turns out to be positive, this means that event number two happened after event number one in his frame, in other words, the light emitted from the back end of Anna's plane arrives first to Bob's eyes.

Given:

dt' = 40ns

dt = 0

dx' = 60

Solution:

$dt' = \sqrt{\frac{1}{(1 - v^2/c^2) } } * (\frac{v*dx'}{c^2} )\\\\\sqrt{1 - \frac {v^2}{c^2 } } = (\frac{v*dx'}{c^2*dt} )\\\\v^2 = \frac{c^4*dt^2}{c^2*dt^2 + dx'^2}\\\\v = \frac{c^2*dt}{\sqrt{c^2*dt^2 + dx'^2} }\\\\v = \frac{(3*10^8)^2*40*10^9}{\sqrt{(3*10^8)^2*(40*10^9)^2 + 60^2} }\\\\v = \frac{3.6*10^9}{61} \\\\v = 0.196 c$

8 0

3 years ago

According to a rule-of-thumb, each five seconds between a lightning flash and the following thunder gives the distance to the fl

jolli1 [7]

<span>There are approximately 1600 meters/mile. Estimating a five second per mile delay in the arrival of the sound, then the estimate of the speed of sound is (a)1600/5 or 320 meters/sec or (B) 320 x 60 sec x 60 min / 1600 m/mile or about 720 miles/hour.</span>

4 0

3 years ago