Equations of motion (EoM) use EoM <span>v2=u2+2ax</span> to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) <span><span>2gH−−−−√</span>=<span><span>2g1625H</span>−−−−−−√</span>+g</span><span> then, solve for H [in terms of g]
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Answer:
5.62 m/s
Explanation:
Newton's law of motion can be used to determine the maximum speed of the elevator. In the question, we are given:
Force exerted by the elevator (R) = 1.7 times the weight of the passenger (m*g)
Thus: R = 1.7*m*g
Distance (s) = 2.3 m
Newton's second law of motion: R - m*g = m*a
1.7*m*g - m*g = m*a
a = 0.7*m*g/m = 0.7*g = 0.7*9.8 = 6.86 m/s²
To determine the maximum speed:
![v^{2} _{f} = v^{2} _{i} + 2as](https://tex.z-dn.net/?f=v%5E%7B2%7D%20_%7Bf%7D%20%3D%20v%5E%7B2%7D%20_%7Bi%7D%20%2B%202as)
![= 0 + 2(6.86)(2.3) = 31.556](https://tex.z-dn.net/?f=%3D%200%20%2B%202%286.86%29%282.3%29%20%3D%2031.556)
![v_{f} = \sqrt{31.556} = 5.62 m/s](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3D%20%5Csqrt%7B31.556%7D%20%3D%205.62%20m%2Fs)
Therefore, the elevator maximum speed is equivalent to 5.62 m/s.
Answer:
3.485e+6 inches. hope this helps
The amplitude of the wave on the given sinusoidal wave graph is 10 cm.
<h3>
What is amplitude of wave?</h3>
The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.
Amplitude of the wave is calculated as follows;
From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.
Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.
Learn more about wave amplitude here: brainly.com/question/25699025
The approximate total distance Earth has traveled since its birth is 3.8×10²¹m.
The Earth moves in a circular orbit of radius <em>r</em> around the Sun and it takes 1 year to go once around the Sun. Thus in a year it travels a distance equal to <em>2πr, </em>which is the circumference of the Earth's orbit.
Therefore in <em>n</em> years that have elapsed, it would travel a distance <em>d</em> given by,
![d=2\pi rn](https://tex.z-dn.net/?f=d%3D2%5Cpi%20rn)
Substitute 3.14 for <em>π, </em>1.5×10¹¹m for <em>r</em> and 4×10⁹ for n.
![d=2\pi rn\\= 2(3.14)(1.5*10^1^1m)(4*10^9)\\ =3.768*10^2^1m=3.8*10^2^1m](https://tex.z-dn.net/?f=d%3D2%5Cpi%20rn%5C%5C%3D%202%283.14%29%281.5%2A10%5E1%5E1m%29%284%2A10%5E9%29%5C%5C%20%3D3.768%2A10%5E2%5E1m%3D3.8%2A10%5E2%5E1m)
Thus, the approximate total distance Earth has traveled since its birth is 3.8×10²¹m.