Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V
