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Zepler [3.9K]
3 years ago
6

The coefficient of friction is a ________ number that represents the resistance to sliding between two __________ in contact wit

h one another.
Physics
2 answers:
pickupchik [31]3 years ago
7 0

APEX

unitless,

surfaces


Varvara68 [4.7K]3 years ago
5 0

Answer:

unitless--surfaces APEX

Explanation:

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A man pushes a shopping cart across a level floor. What force resists the effort force? A) gravity B) friction C) the normal for
Kipish [7]

Answer:

B) Friction

Explanation:

Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.

For an object sliding on a flat surface, the force of friction has magnitude:

F_f = \mu_k mg

where

\mu_k is the coefficient of kinetic friction

m is the mass of the object

g is the acceleration of gravity

The direction of the force of friction is always opposite to the direction of motion of the object.

In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is

F_f = \mu_s mg

where \mu_s is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.

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3 years ago
In the following atomic model, where does the strong nuclear force happen? mc006-1.jpg outside A between A and B between B and C
lukranit [14]
The answer is the third option
3 0
3 years ago
State examples of a transverse wave. ​
laiz [17]

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

6 0
3 years ago
Read 2 more answers
A monochromatic red laser beam emitting 1 mW at a wavelength of 638 nm is incident on a silicon solar cell. Find the following:
tia_tia [17]

Answer:

Explanation:

First we calculate the energy of the photon

E=(Planck constant × speed of light in vacuum)÷ wave length

E=\frac{6.626*10^{34}*2.998*10^{8}  }{638*10^{-9} } = 3.114*10^{49}

Next we find the total energy per second

total energy= 1*10^{-3}W *\frac{1JS^{-1} }{1W}  = 1*10^{-3} JS^{-1}

Next we calculate the number the photon per second

= total energy ÷ energy of 1 photon

= \frac{1*10^{-3} JS^{-1}  }{3.114*10^{49} } =  3.21*10^{-53}  \ photons/sec

8 0
3 years ago
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