Then the magnitude of the net force is the difference between the two forces,
and its direction is the same as the direction of the greater one.
Answer:
Distance, d = 112.5 meters
Explanation:
Initially, the bicyclist is at rest, u = 0
Final speed of the bicyclist, v = 30 m/s
Acceleration of the bicycle, 
Let s is the distance travelled by the bicyclist. The third equation of motion is given as :
s = 112.5 meters
So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.
Answer:
2 m/s^2
Explanation:
a = v^2/r
a = (10m/s)^2 / 50m
a = 2 m/s^2
Leave a like and mark brainliest if this helped
Leave a like and mark brainliest if this helped
C.
Newton’s Second Law is F=ma (force is equal to the mass multiplied by acceleration), however, the equation can be rearranged to isolate and calculate mass from force over acceleration. Therefore, m=F/a
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span>
<span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span>
<span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)=<span>
<span>1960.74 rev/min
</span></span>
