Answer:
n = c/v = (3.00 x 108 m/s)/(2.76 x 108 m/s) = 1.09. This does not equal any of the indices of refraction listed in the table.
Answer:
65 m/s
Explanation:
v=v0+at <=> v = 11 + 12 t ∧ t = 4.5 s <=> v = 11 + 12×4.5 <=> v = 65 m/s
Answer:
The deceleration is ![a = - 76.27 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%20-%2076.27%20m%2Fs%5E2)
Explanation:
From the question we are told that
The height above firefighter safety net is ![H = 14 \ m](https://tex.z-dn.net/?f=H%20%20%3D%2014%20%5C%20m)
The length by which the net is stretched is ![s = 1.8 \ m](https://tex.z-dn.net/?f=s%20%3D%20%201.8%20%5C%20m)
From the law of energy conservation
![KE_T + PE_T = KE_B + PE_B](https://tex.z-dn.net/?f=KE_T%20%2B%20PE_T%20%3D%20%20KE_B%20%2B%20PE_B)
Where
is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )
and
is the potential energy of the before jumping which is mathematically represented at
![PE_T = mg H](https://tex.z-dn.net/?f=PE_T%20%20%3D%20mg%20H)
and
is the kinetic energy of the person just before landing on the safety net which is mathematically represented at
![KE_B = \frac{1}{2} m v^2](https://tex.z-dn.net/?f=KE_B%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2)
and
is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )
So the above equation becomes
![mgH = \frac{1}{2} m v^2](https://tex.z-dn.net/?f=mgH%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2)
=> ![v = \sqrt{2 gH }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Csqrt%7B2%20gH%20%7D)
substituting values
![v = 16.57 m/s](https://tex.z-dn.net/?f=v%20%3D%20%2016.57%20m%2Fs)
Applying the equation o motion
![v_f = v + 2 a s](https://tex.z-dn.net/?f=v_f%20%3D%20%20v%20%20%2B%202%20a%20s)
Now the final velocity is zero because the person comes to rest
So
![0 = 16.57 + 2 * a * 1.8](https://tex.z-dn.net/?f=0%20%3D%2016.57%20%2B%202%20%2A%20a%20%2A%201.8)
![a = - \frac{16.57^2 }{2 * 1.8}](https://tex.z-dn.net/?f=a%20%3D%20%20-%20%5Cfrac%7B16.57%5E2%20%7D%7B2%20%2A%201.8%7D)
![a = - 76.27 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%20-%2076.27%20m%2Fs%5E2)
The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop.
Hope this helps!