Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
It’s c.) I think so at least
<em><u>The</u></em><em><u> </u></em><em><u>atomic</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>consists</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>protons</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u>.</u></em>
<em><u>Additional</u></em><em><u> </u></em><em><u>information</u></em><em><u>:</u></em>
<em><u>Protons</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>positive</u></em><em><u>ly</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particl</u></em><em><u>e</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>negative</u></em><em><u>ly</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u>.</u></em>
<em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>:</u></em><em><u>)</u></em>
Answer:
-2040 m/s²
Explanation:
Taking toward the wall to be positive, the initial velocity is 10.1 m/s and the final velocity is -8.3426 m/s.
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (-8.3426 m/s − 10.1 m/s) / 0.00905 s
a = -2040 m/s²
Answer: 96N
Explanation:
To calculate the velocity of the impact On the persons head, we have
h = gt²/2
14 = 9.81t²/2
t² = 28/9.8
t² = 2.86
t = 1.69s
V = u + at
V = 0 + 9.81*1.69
V = 16.58m/s
a(average) = (v1² + v2²) /2Δy
a(average) = 16.58² + 0)/2 * 0.005
a(average) = 274.8964/0.01
a(average) = 27489.64m/s²
Using newton's second law of motion,
F(average) = m * a(average)
F(average) = 0.0035 * 27489.64
F(average) = 96.21N
Therefore the force needed by the acorn to do much damage starts from 96N