Answer:
U = 0.413 J
Explanation:
the potential energy between two charges q1 and q2 is given by the following formula:
(1)
k: Coulomb's constant = 8.98*10^9 NM^2/C^2
q1: first charge = 4.6 μC = 4.6*10^-6 C
q2: second charge = 1.0 μC*10^-6 C
r: distance between charges = 10.0 cm = 0.10 m
You replace the values of all variables in the equation (1):
Hence, the energy between charges is 0.413 J
Answer:
50kg.m/s
Explanation:
In order to find momentum you must use the formula P=mv
p= momentum
m=mass
v= velocity
so in other words, momentum= mass times velocity
or in this case, momentum= 10 times 5 :)
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
By Boyle's law:
P₁V₁ = P₂V₂
70*8 = P<span>₂*4
</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
</span>
P<span>₂ = 140 kiloPascals.</span>
