Answer:
a) 31.4 m/s
b) 50.2 m
Explanation:
a) When an object is free falling, its speed is determined by the gravity force giving it acceleration. Equation for the velocity of free fall started from the rest is:
v = g • t
g - is gravitational acceleration which is 9.81 m/s^2, sometimes rounded to 10
t - is the time of free fall
So:
v = 9.81 m/s^2 • 3.2
v = 31.4 m/s ( if g is rounded to 10, then the velocity is 10 • 3.2 = 32 m/s)
b) To determine the distance crossed in free fall we use the equation:
s = v0 + gt^2/2
v0 - is the starting velocity (since object started fall from rest, its v0 is 0)
s = gt^2/2
s = 9.81 m/s^2 • 3.2^2 / 2
s = 50.2 m (if we round g to 10 then the distance is 10 • 3.2^2/2 = 51.2 meters)
It is E=something which leases another something equaling another something
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie
