Answer:
a=2.304×10¹⁶m/s²
Explanation:
Given data
Distance d=2.5 nm=2,5×10⁻⁹m
Mass of proton m=1.6×10⁻²⁷kg
charge of proton q=1.6×10⁻¹⁹C
To find
acceleration a
Solution
Apply the Coulombs Law
Where k is coulombs constant (k=9×10⁹Nm²/C²)
q=q₁=q₂
r=d
So
Answer:
a) # buses = 7
Explanation:
For this exercise we use the kinematic equations, let's find the time it takes to reach the same height
y = t - ½ g t²
Let's decompose the speed, with trigonometry
v₀ₓ = v₀ cos θ
= v₀ sin θ
v₀ₓ = 40 cos 32
v₀ₓ = 33.9 m / s
= 40 sin32
= 21.2 m / s
When it arrives it is at the same initial height y = 0
0 = ( - ½ gt) t
That has two solutions
t = 0 when it comes out
t = 2 / g when it arrives
t = 2 21.2 /9.8
t = 4,326 s
We use the horizontal displacement equation
x = vox t
x = 33.9 4.326
x = 146.7 m
To find the number of buses we can use a direct proportions rule
# buses = 146.7 / 20
# buses = 7.3
# buses = 7
The distance of the seven buses is
L = 20 * 7 = 140 m
b) let's look for the scope for this jump
R = vo2 sin2T / g
R = 40 2 without 2 32 /9.8
R = 146.7 m
As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)