Answer:
The speed of proton when it emerges through the hole in the positive plate is .
Explanation:
Given that,
A parallel-plate capacitor is held at a potential difference of 250 V.
A A proton is fired toward a small hole in the negative plate with a speed of,
We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :
So, the speed of proton when it emerges through the hole in the positive plate is .
Answer:
λ = 396.7 nm
Explanation:
For this exercise we use the diffraction ratio of a grating
d sin θ = m λ
in general the networks works in the first order m = 1
we can use trigonometry, remembering that in diffraction experiments the angles are small
tan θ = y / L
tan θ = = sin θ
sin θ = y / L
we substitute
= m λ
with the initial data we look for the distance between the lines
d =
d = 1 656 10⁻⁹ 1.00 / 0.600
d = 1.09 10⁻⁶ m
for the unknown lamp we look for the wavelength
λ = d y / L m
λ = 1.09 10⁻⁶ 0.364 / 1.00 1
λ = 3.9676 10⁻⁷ m
λ = 3.967 10⁻⁷ m
we reduce nm
λ = 396.7 nm
Answer:
t = 106π / 30*2.1
Explanation:
=> 106
=> 106 x 2π/60
=> 106/30π
∝ = -2.1 rad/sec²
=> 0
=
+ ∝t
∴ ( -
) / ∝ = t
t = 106π / 30*2.1