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tester [92]
3 years ago
13

3. If you wanted the athletic shoe to slide faster on a surface, what might you do to the shoe? Provide at least two ways to mak

e the shoe slide faster, and explain in terms of the force of friction and the coefficient of friction.
Physics
1 answer:
lesya692 [45]3 years ago
6 0
In order to make the athletic shoe to slide faster of a given surface you had to reduce the friction coefficient of the shoe with the surface.

One way to do that is to put some grease of the sole of the shoe, which undoubtedly would reduce the friction coefficient drastically and the shoe would slide more easily.

Also, if you make the sole smoother the shoe the friction coefficient would decrease and, again, the show would slide mor easily.
You might be interested in
Light is shone on a diffraction grating
Pani-rosa [81]

Answer:

    λ = 482.05 nm

Explanation:

The diffraction phenomenon and the diffraction grating is described by the expression

         d sin θ = m λ

where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction

in this case they indicate the distance between slits, the angle and the order of diffraction

         λ = \frac{d sin \theta }{m}d sin θ / m

let's calculate

         λ = 1.00 10⁻⁶ sin 74.6 / 2

         λ = 4.82048 10⁻⁷ m

Let's reduce to nm

         λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)

         λ = 482.05 nm

3 0
3 years ago
A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t
vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

\mu =7.20 g/m = 0.0072 kg/m is the mass linear density

Solving the equation,

f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

f=\frac{v}{2L}

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m

4 0
4 years ago
Un vehicle surt, en un instant, d'un punt A cap a un altre punt B amb una velocitat constant de 36 km/h. Després d'un minut(60 s
kati45 [8]



Ask Siri or goggle has all answers



6 0
3 years ago
The earth orbits a start called the ____?
Aleks [24]

Answer:

the earth orbits a star called the sun

6 0
3 years ago
A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1
kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

\mu = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s

Frequency is given by

f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz

Angular frequency is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s

Maximum velocity of a particle is given by

v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0
3 years ago
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