I have no idea how too do that work sheet but try looking for the answer key online by searching for the name of the worksheet
Answer:
(a) ![E=2.42eV](https://tex.z-dn.net/?f=E%3D2.42eV)
(b) ![E=1.45*10^{-5}eV](https://tex.z-dn.net/?f=E%3D1.45%2A10%5E%7B-5%7DeV)
(c) ![E=1.66*10^{-7}eV](https://tex.z-dn.net/?f=E%3D1.66%2A10%5E%7B-7%7DeV)
Explanation:
The Planck-Einstein relation allows us to know the energy (E) of a photon, knowing its frequency (f). According to this relation, the energy of the photon is defined as:
![E=hf](https://tex.z-dn.net/?f=E%3Dhf)
Here h is the Planck constant.
(a)
![E=(4.14*10^{-15}eV\cdot s)(585*10^{12}Hz)\\E=2.42eV](https://tex.z-dn.net/?f=E%3D%284.14%2A10%5E%7B-15%7DeV%5Ccdot%20s%29%28585%2A10%5E%7B12%7DHz%29%5C%5CE%3D2.42eV)
(b)
![E=(4.14*10^{-15}eV\cdot s)(3.50*10^{9}Hz)\\E=1.45*10^{-5}eV](https://tex.z-dn.net/?f=E%3D%284.14%2A10%5E%7B-15%7DeV%5Ccdot%20s%29%283.50%2A10%5E%7B9%7DHz%29%5C%5CE%3D1.45%2A10%5E%7B-5%7DeV)
(c)
![E=(4.14*10^{-15}eV\cdot s)(40.0*10^{6}Hz)\\E=1.66*10^{-7}eV](https://tex.z-dn.net/?f=E%3D%284.14%2A10%5E%7B-15%7DeV%5Ccdot%20s%29%2840.0%2A10%5E%7B6%7DHz%29%5C%5CE%3D1.66%2A10%5E%7B-7%7DeV)
<span>The amplitude because that controls the height of the wave. Correct answer: Amplitude.</span>
Answer:
a
t = 1.235 s
b
![h_{max} = 2.55 \ m](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%20%202.55%20%5C%20m)
c
![v = -7.10 m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%20-7.10%20m%2Fs)
Explanation:
From the question we are told that
The initial velocity is ![u = 5.0 \ m/s](https://tex.z-dn.net/?f=u%20%20%3D%205.0%20%5C%20%20m%2Fs)
The take of height is ![h = -1.30 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%20-1.30%20%5C%20%20m)
The negative sign show that the height is on the negative y-axis when the take off point is consider as the origin
Generally from the kinematic equation we have that
![s = ut + \frac{1}{2}at^2](https://tex.z-dn.net/?f=s%20%3D%20%20ut%20%2B%20%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
=> ![-1.30 = 5.0 t + \frac{1}{2} (-9.8)t^2](https://tex.z-dn.net/?f=-1.30%20%3D%20%205.0%20t%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%28-9.8%29t%5E2)
Here g is negative given that the swimmers jump motion is against gravity
![4.9t^2 - 5t - 1.30](https://tex.z-dn.net/?f=4.9t%5E2%20-%20%205t%20%20-%20%201.30)
solving using quadratic formula we obtain that
t = 1.235 s
Generally the highest point is mathematically evaluated as
![h_{max} = \frac{v^2 - u^2 }{2 * (-g) }](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%20%20%5Cfrac%7Bv%5E2%20%20-%20%20u%5E2%20%7D%7B2%20%2A%20%28-g%29%20%7D)
Here v = 0 m/s since the velocity at the highest point is 0
![h_{max} = \frac{0^2 - 5^2 }{2 * (-9.8) }](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%20%20%5Cfrac%7B0%5E2%20%20-%20%205%5E2%20%7D%7B2%20%2A%20%28-9.8%29%20%7D)
![h_{max} = 2.55 \ m](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%20%202.55%20%5C%20m)
Generally her velocity when the feet hits the water is mathematically evaluated from kinematic equation as
![v = \pm \sqrt{ u^2 + 2 (-g) * (-h)}](https://tex.z-dn.net/?f=v%20%20%3D%20%20%5Cpm%20%5Csqrt%7B%20u%5E2%20%20%2B%202%20%28-g%29%20%2A%20%20%28-h%29%7D)
![v = -7.10 m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%20-7.10%20m%2Fs)
The negative value of the velocity is selected because the velocity is on the negative y-axis
Answer:
The acceleration is about 9.8 m/s2 (down) when the ball is falling.
Explanation:
The ball at maximum height has velocity zero
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)
![v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%200%3Du-9.8%5Ctimes%20t%5C%5C%5CRightarrow%20u%3D9.8t)
The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero