<span>The Badminton World Federation</span>
Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get
10N = 50 kg · a
A newton is just 
The s^2 and 50 kg you multiply
The kg's cancel and 10/50 is 1/5
So the acceleration is 1/5 m/s^2
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Explanation:
For this exercise let's use the relationship between momentum and momentum.
I = F t = Δp
in this case the final velocity is zero
F t = 0 -m v₀
F = m v₀ / t
in order to answer the question we must assume that the two vehicles have the same mass and speed
concrete barrier
F₁ = -p₀ / 0.1
F₁ = - 10 p₀
barrier collapses
F₂ = -p₀ / 1
let's look for the relationship of the forces
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
