Answer:
D
Explanation:
The friction force is the weight force times the coefficient of friction. So diving by the coefficient gives you the weight force which is equivalent to the normal force.
Answer: 1.77 s
Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:
xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then
(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s
then by solving the quadratric equation in t
t=1.77 s
We're u can never put it back together
Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.?
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m