Answer:
KE = KE (incidental) - KE of emitted photons
or KE = h * f - Wf
So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy
If you double the frequency then h * f = 6.16
KE = 6.16 - 1.2 = 4.96 eV
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for 
DAC:
⇒ 
Now for the BAC:
⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle, 
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 
After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x =
Answer:
A. ls sensitive to heat
this is correct answer
I hope it's helpful for you.......
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m
