Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. car 1 suddenly starts to brake
1 answer:
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Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
Therefore, the height of the cliff is 121.276 m
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F =
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F =
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = 21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles