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
Answer:
30 metres.
Explanation:
Given that a red ball moves horizontally in a 30 m long tube.
Displacement is the distance travelled in a specific direction. It has both magnitude and direction.
Since the motion is horizontal, it moves is a certain direction.
Within the stipulation of time, the displacement will be the distance covered in the horizontal direction which is 30 metres.
Therefore, the displacement of the motion of the red ball is 30 metres.
Write a:
● Hypothesis
●Procedures
●Materials
●Data Table
●Conclusion
●Variables (Independent and dependent)
●Bibliography
Answer:

Explanation:
consider the mass of each train car be m
m₁ = m₂ = m₃ = m
speed of the three identical train
u₁ = u₂ = u₃ = 1.8 m/s
m₄ = m u₄ = 4.5 m/s
m₅ = m u₅ = 0 (initial velocity )
final velocity
v₁ = v₂ = v₃ = v₄ = v₅ = v
using conservation of momentum
m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅
m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

