Answer:
49 N
Explanation:
In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,
F = ma
that the net force acting on the box, F, must be zero as well.
Here there are two forces acting on the box in the horizontal direction while it is moving:
- The force of push applied by the guy, F
- The frictional force, 
For an object moving on a flat surface, the frictional force is given by
where
is the coefficient of friction
m is the mass of the box
g is the acceleration of gravity
So the equation of the forces becomes
And substituting:
We find the force that must be applied by the guy:
According to the research, the correct option the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.
<h3>What is the
eardrum?</h3>
It is the membrane found in the middle ear of vertebrate animals, separating this sector from the external auditory canal.
When sound waves enter through the external auditory canal, the eardrum vibrates, transmitting its movement to the middle ear through a series of bones and in this way, the pressure change is transformed into a mechanical movement.
Therefore, we can conclude that according to the research, the correct option is the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.
Learn more about the eardrum here: brainly.com/question/12770491
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Mechanical and Chemical. (Weathering and erosion)
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
