Ok so the formula is d=vi(t)+½at² and when you substitute it you should get 172.5meters
The unit used to measure wavelength is a Nano-meter
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:
3.066×10^21 photons/(s.m^2)
Explanation:
The power per area is:
Power/A = (# of photons /t /A)×(energy / photon)
E/photons = h×c/(λ)
photons /t /A = (Power/A)×λ /(h×c)
photons /t /A = (P/A)×λ/(hc)
photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)
= 3.066×10^21
Therefore, the number of photons per second per square meter 3.066×10^21 photons/(s.m^2).