To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
![a = \frac{F}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF%7D%7Bm%7D)
![a = \frac{11*10^3}{80}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B11%2A10%5E3%7D%7B80%7D)
![a = 137.5m/s](https://tex.z-dn.net/?f=a%20%3D%20137.5m%2Fs)
From the cinematic equations of motion we know that
![v_f^2-v_i^2 = 2ax](https://tex.z-dn.net/?f=v_f%5E2-v_i%5E2%20%3D%202ax)
Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
![v_f^2-v_i^2 = 2ax](https://tex.z-dn.net/?f=v_f%5E2-v_i%5E2%20%3D%202ax)
![0-v_i^2 = 2(-g)x](https://tex.z-dn.net/?f=0-v_i%5E2%20%3D%202%28-g%29x)
![v_i =\sqrt{2gx}](https://tex.z-dn.net/?f=v_i%20%3D%5Csqrt%7B2gx%7D)
![v_i = \sqrt{2*9.8*4.8}](https://tex.z-dn.net/?f=v_i%20%3D%20%5Csqrt%7B2%2A9.8%2A4.8%7D)
![v_i = 9.69m/s](https://tex.z-dn.net/?f=v_i%20%3D%209.69m%2Fs)
From the equation of motion where acceleration is equal to the velocity in function of time we have
![a = \frac{v_i}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv_i%7D%7Bt%7D)
![t = \frac{v_i}{a}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bv_i%7D%7Ba%7D)
![t =\frac{9.69}{137.5}](https://tex.z-dn.net/?f=t%20%3D%5Cfrac%7B9.69%7D%7B137.5%7D)
![t = 0.0705s](https://tex.z-dn.net/?f=t%20%3D%200.0705s)
Therefore the time required is 0.0705s
Depends on what type of mirror that is. I am going to assume this is a plain mirror (from the phrase), which means the height and width of the object and image is exactly the same.
Answer:
26.833 N
Explanation:
The computation of the resaltant of two forces is shown below:
Given that
Force A = 12N
Force B = 24N
Based on the above information
Resultant R is
![=\sqrt{A^2 + B^2 + 2AB \times cos \theta}\\\\=\sqrt{144 + 576 + 2\times 24\times 12\times cos90^{\circ}}\\\\=\sqrt{144+576+576\times 0}\\\\=\sqrt{720}](https://tex.z-dn.net/?f=%3D%5Csqrt%7BA%5E2%20%2B%20B%5E2%20%2B%202AB%20%5Ctimes%20cos%20%5Ctheta%7D%5C%5C%5C%5C%3D%5Csqrt%7B144%20%2B%20576%20%2B%202%5Ctimes%2024%5Ctimes%2012%5Ctimes%20cos90%5E%7B%5Ccirc%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B144%2B576%2B576%5Ctimes%200%7D%5C%5C%5C%5C%3D%5Csqrt%7B720%7D)
=26.833 N
Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>
<u></u>
<h3>What is vertical acceleration?</h3>
A vertical acceleration is typically one for which the direction of the vector is vertically upward, usually aligned with and opposite to the gravity vector. But this is a descriptive term, not a rigorous or technical term. A car may accelerate along a road and that would generally be assumed to be a horizontal.
The vector perpendicular to this direction, as perhaps a suspension motion over a bump, would be described as vertical even if it is not strictly vertical.
Note that acceleration is defined as the rate of change of the velocity vector. But the gravitation vector, ‘g’, generally vertically downward, is often denoted by what acceleration a mass in free fall (absent air resistance) would experience, i.e. the relationship between mass and weight.
Learn more about vertical acceleration
brainly.com/question/19528199
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