Here it is an application of Newton's III law
as we know by Newton's III law that every action has equal and opposite reaction
So here as we know that two boys jumps off the boat with different forces
from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude
So boat must have an opposite force on front end with magnitude 45 N
Now similar way we can say
from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude
So boat must have an opposite force on front end with magnitude 60 N
So here net force due to both jump on the boat is given by
so boat will have net force F = 15 N in forward direction due to both jumps
Another name for these two words is "constant" and you want to have a "constant", because you want something to compare your experimental group to, to see whether data had changed or not. So you have placebos or a double- blind to compare your experimental group to it and also so you know you don't have a bias or anything in the study.
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
I think the answer is 4) All of the above!! :)
