When looking at it from a physics standpoint, the thick padding decreases the amount of force that a player experiences. Force equals mass times acceleration where acceleration is distance over time. The thick padding increases the amount of time the player makes contact with the post which decreases force.
Answer:
400 J
Explanation:
Given:
Δy = 4.00 m
v₀ = 0 m/s
a = 5.00 m/s²
Find: v²
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (5.00 m/s²) (4.00 m)
v² = 40.0 m²/s²
Find KE:
KE = ½ mv²
KE = ½ (20.0 kg) (40.0 m²/s²)
KE = 400 J
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA =
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0
momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ
Solve the two equations for v₂ and θ.
The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
They are asking you about the direction so you have to use trigonometry, finding that the direction is Ф=arctan(3.8/12)=17.57° north of east.