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3 years ago

Is the amount of friction greater between rough surfaces or smooth surfaces? Why?

Physics

1 answer:

nalin [4]3 years ago

6 0

Answer:

friction occurs because no surface is perfectly smooth.

Explanation:

rougher surfaces have more friction between them.

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6. Which of these contain muscles that are not under the mind's control? (Select all that apply.)

Arada [10]

Answer:

a c

Explanation:

edu 2021

3 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

3 years ago

Helppppppppppppppp asappppppppppppp

vichka [17]

The answer to the problem b.

8 0

3 years ago

The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this is powers of ten i

DerKrebs [107]

1.4 times 10 in power of 10

8 0

3 years ago

A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc

hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

$E V = log_2(\dfrac{N^2}{t})$

N is the diameter of the aperture and t is the time of exposure

now,

$log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})$

$\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}$

inserting all the values

$\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}$

N₂² = 4

N₂ = 2 mm

hence , the correct answer is option E

4 0

3 years ago