Answer:
For 6.0 eV
0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s
For 600 eV
1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s
Explanation:
See attachment for calculation
The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.
(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m
Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.
(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ
θ=37.69 degrees
Hopes this helps!
P.S. I hope this is right. If not sorry in advance.
<span>Answer:
Well, let's start by finding the pressure due to the "extra" height of the mercury.
p = 1.36e4 kg/m³ · (0.105m - 0.05m) · 9.8m/s² = 7330 N/m² = 7330 Pa
The pressure at B is clearly p_b = p_atmos = p_gas + 7330 Pa
The pressure at A is p_a = p_gas = p_atmos - 7330 Pa
c) 1 atm = 101 325 Pa
Then p_gas = 101325 Pa - 7330 Pa = 93 995 Pa</span>
Answer: I don’t know Kiane I’m looking for the anwser too—Gianny
Explanation:
