<span>Answer:
Fg,y = mg(cos q1)
Fg,x = mg(sin q1)
Fapplied,x = Fapplied(cos q2)
Fapplied,y = Fapplied(sin q2)
Fn = Fg,y â’ Fapplied,y
Fk =mkFn
Fapplied,x â’ Fk â’ Fg,x
ax=fapplied-fk-fg,x/ m</span>
Answer:
wait what do u mean by healthy beccause i choose C
Explanation:
Por the impulse, use the formula:
I = F * t
Now, we know:
4000 N = 4000 kg*m/s²
Replacing we have:
I = 4000 kg*m/s² * 0,200 s
Resolving:
I = 800 kg.m/s²
The impulse is <u>800 kg.m/s²</u>
Answer:
9.89 m/s.
Explanation:
Given that,
The radius of the circular arc, r = 25 m
The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²
Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

So, the maximum speed of the car should be 9.89 m/s.