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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.
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Why we don’t feel atmospheric pressure?
Veseljchak [2.6K]

Answer:

The reason we can't feel it is that the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards, so there's no pressure difference and no need for us to exert any effort.

8 0
3 years ago
Read 2 more answers
A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 n
maksim [4K]

Answer:

A) receding from the earth

B) 3.078x10^6m/s

Explanation:

  • A) receding from the earth

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

  • B) 3.078x10^6m/s

to calculate the relative speed we use the following formula:

v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )

where c is the speed of light: c=3x10^8m/s

\lambda_{1} is the wavelength emited by the source, and

\lambda_{2} is the wavelength measured on earth.

we substitute all the values and do the calculations:

v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s

the relative speed is: 3.078x10^6m/s

5 0
3 years ago
Define refractive index.​
levacccp [35]
The refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.

the ratio of the velocity of light in a vacuum to its velocity in a specified medium.
8 0
2 years ago
Read 2 more answers
A 7.0-μC point charge and a point charge are initially extremely far apart. How much work does it take to bring the point charge
vazorg [7]

Answer:1.008 ×10^-14/rJ

Where r is the distance from.which the charge was moved through.

Explanation:

From coloumbs law

Work done =KQq/r

Where K=9×10^9

Q=7×10^-6C

q=e=1.6×10^-19C

Micro is 10^-6

W=9×10^9×7×10^-6×1.6×10^-19/r=100.8×10^-16/r=1.008×10^-14/rJ

r represent the distance through which the force was used to moved the charge through.

4 0
3 years ago
Read 2 more answers
If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel
expeople1 [14]

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

V = u + at

Substituting into the equation, we have;

V = 0 + 9.8 * 3.6

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Substituting the values into the equation;

S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}

S = 0 + 4.9*12.86

S = 0.5 *36

S = 63.50m

Therefore, the height of the tower is 63.50m.

6 0
3 years ago
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