Answer: d)
Explanation: In order to justify the answer we have to consider that the energy of photons directely depent on the frequency so the energy is inverselly dependent of the wavelegth.
If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as b) since it is not posible since UV photon have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:
c=λν c is the speed of light, λ the wavelegth and ν the frequency.
d) Certainly will be more more IR photons than UV photons to get the same beam power.
Answer:
so his far point according to this pair of glass is 200 cm
Explanation:
power of old pair of corrective glasses is given as

now we have



now we know that for normal vision the maximum distance of vision is for infinite distance
so by lens formula we have



so his far point according to this pair of glass is 200 cm
Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body


P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s
<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890.
For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time.
Add the two times together for the total.
The alternative is to calculate the initial and final velocity so that you have more information to work with.
Answer:
The angular speed of the object is 0.0281 rad/s
The linear speed of the object is 0.169 ft/s
Explanation:
Given;
radius of the circle, r = 6 ft
time of motion of the object around the circle, t = 80 s
central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad
The angular speed of the object is calculated as;

The linear speed of the object is calculated as;
v = ωr
v = 0.0281 rad/s x 6ft
v = 0.169 ft/s