Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = 
mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = 
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p = 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = 
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = 
v/c= 2.33 10⁻⁴
12. The answer would be C. 1.50 s. This is because if you divide 60 by 40, you will get 1.5.
13. For this one I'm not sure, but what I can tell you is that the heavier something is the faster it will sink, the lighter it is, it will float.
Set up a proportion
1 mile/1.6 km = 20,000miles/x
Cross Multiply
x = (20,000) * (1.6)
x = 32,000 kilometers
Answer: 0.642mm
Explanation: F= force = 5.2×10^-16 N,
v = velocity of electron = 1.2×10^7 m/s,
m = mass of electron = 9.11×10^-31 kg.
We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.
Recall that f = ma.
Where a = acceleration
This acceleration of vertical because it occurred when the object deflected.
5.2×10^-16 = 9.11×10^-31 (ay)
ay = 5.2×10^-16 / 9.11×10^-31
ay = 5.71×10^14 m/s²
For the horizontal motion, x = vt
Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,
By substituting the parameters, we have that
0.019 = 1.27×10^7 × t
t = 0.019 / 1.27 × 10^7
t = 1.5×10^-9 s
The vertical distance (y) is gotten by using the formulae below
y = ut + at²/2
but u = 0
y = at²/2
y = 5.71×10^14 × (1.5×10^-9)²/2
y = 0.00128475/2
y = 0.000642m = 0.642mm
