Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking
Compression and rarefaction are two phenomenon occurs in longitudunal wave!
when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !
Answer:
a) ![h=250\ m](https://tex.z-dn.net/?f=h%3D250%5C%20m)
b) ![\Delta h=0.0835\ m](https://tex.z-dn.net/?f=%5CDelta%20h%3D0.0835%5C%20m)
Explanation:
Given:
- upward acceleration of the helicopter,
![a=5\ m.s^{-2}](https://tex.z-dn.net/?f=a%3D5%5C%20m.s%5E%7B-2%7D)
- time after the takeoff after which the engine is shut off,
![t_a=10\ s](https://tex.z-dn.net/?f=t_a%3D10%5C%20s)
a)
<u>Maximum height reached by the helicopter:</u>
using the equation of motion,
![h=u.t+\frac{1}{2} a.t^2](https://tex.z-dn.net/?f=h%3Du.t%2B%5Cfrac%7B1%7D%7B2%7D%20a.t%5E2)
where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation
![h=0+0.5\times 5\times 10^2](https://tex.z-dn.net/?f=h%3D0%2B0.5%5Ctimes%205%5Ctimes%2010%5E2)
![h=250\ m](https://tex.z-dn.net/?f=h%3D250%5C%20m)
b)
- time after which Austin Powers deploys parachute(time of free fall),
![t_f=7\ s](https://tex.z-dn.net/?f=t_f%3D7%5C%20s)
- acceleration after deploying the parachute,
![a_p=2\ m.s^{-2}](https://tex.z-dn.net/?f=a_p%3D2%5C%20m.s%5E%7B-2%7D)
<u>height fallen freely by Austin:</u>
![h_f=u.t_f+\frac{1}{2} g.t_f^2](https://tex.z-dn.net/?f=h_f%3Du.t_f%2B%5Cfrac%7B1%7D%7B2%7D%20g.t_f%5E2)
where:
initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
time of free fall
![h_f=0+0.5\times 9.8\times 7^2](https://tex.z-dn.net/?f=h_f%3D0%2B0.5%5Ctimes%209.8%5Ctimes%207%5E2)
![h_f=240.1\ m](https://tex.z-dn.net/?f=h_f%3D240.1%5C%20m)
<u>Velocity just before opening the parachute:</u>
![v_f=u+g.t_f](https://tex.z-dn.net/?f=v_f%3Du%2Bg.t_f)
![v_f=0+9.8\times 7](https://tex.z-dn.net/?f=v_f%3D0%2B9.8%5Ctimes%207)
![v_f=68.6\ m.s^{-1}](https://tex.z-dn.net/?f=v_f%3D68.6%5C%20m.s%5E%7B-1%7D)
<u>Time taken by the helicopter to fall:</u>
![h=u.t_h+\frac{1}{2} g.t_h^2](https://tex.z-dn.net/?f=h%3Du.t_h%2B%5Cfrac%7B1%7D%7B2%7D%20g.t_h%5E2)
where:
initial velocity of the helicopter just before it begins falling freely = 0
time taken by the helicopter to fall on ground
height from where it falls = 250 m
now,
![250=0+0.5\times 9.8\times t_h^2](https://tex.z-dn.net/?f=250%3D0%2B0.5%5Ctimes%209.8%5Ctimes%20t_h%5E2)
![t_h=7.1429\ s](https://tex.z-dn.net/?f=t_h%3D7.1429%5C%20s)
From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
<u>remaining time,</u>
![t'=t_h-t_f](https://tex.z-dn.net/?f=t%27%3Dt_h-t_f)
![t'=7.1428-7](https://tex.z-dn.net/?f=t%27%3D7.1428-7)
![t'=0.1428\ s](https://tex.z-dn.net/?f=t%27%3D0.1428%5C%20s)
<u>Now the height fallen in the remaining time using parachute:</u>
![h'=v_f.t'+\frac{1}{2} a_p.t'^2](https://tex.z-dn.net/?f=h%27%3Dv_f.t%27%2B%5Cfrac%7B1%7D%7B2%7D%20a_p.t%27%5E2)
![h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2](https://tex.z-dn.net/?f=h%27%3D68.6%5Ctimes%200.1428%2B0.5%5Ctimes%202%5Ctimes%200.1428%5E2)
![h'=9.8165\ m](https://tex.z-dn.net/?f=h%27%3D9.8165%5C%20m)
<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>
![\Delta h=h-(h_f+h')](https://tex.z-dn.net/?f=%5CDelta%20h%3Dh-%28h_f%2Bh%27%29)
![\Delta h=250-(240.1+9.8165)](https://tex.z-dn.net/?f=%5CDelta%20h%3D250-%28240.1%2B9.8165%29)
![\Delta h=0.0835\ m](https://tex.z-dn.net/?f=%5CDelta%20h%3D0.0835%5C%20m)
Answer:Same magnitude
Explanation:
When ball is dropped from shoulder height h then velocity at the bottom is given by
![v_1=\sqrt{2gh}](https://tex.z-dn.net/?f=v_1%3D%5Csqrt%7B2gh%7D)
if it makes elastic collision then it will acquire the same velocity and riser up to the same height
If m is the mass of ball then impulse imparted is given by
![J=m(v_2-v_1)](https://tex.z-dn.net/?f=J%3Dm%28v_2-v_1%29)
![J=2m\sqrt{2gh}](https://tex.z-dn.net/?f=J%3D2m%5Csqrt%7B2gh%7D)
Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.