Amy crashed her bike into the fence. She was thrown over it onto the lawn. Which Newton law applies
1 answer:
You might be interested in
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call
the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
(1)
where
is the mass of ball A
is the mass of ball B
is the initial velocity of ball A
is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)
If we solve (1) to find, we find that the final velocity of the balls is
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
is the final velocity of ball A
is the final velocity of ball B
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
If we call
where
If we solve (1) to find
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h =
h =
h = 1.55 m