I’ve done this before the answer is B
Answer:
94
Explanation:
f = 2.57 x 10^13 Hz
E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J
Energy of each photon = h f
Where, h is Plank's constant
Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J
Number of photons = Total energy / energy of one photon
N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement
putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement =
⇒
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m