I would say B. Because actual mass would ricochet off the sidewalk.
Answer: when a object gets slowed down, it's force is either going into friction and drag, if it's on the ground, and weight+drag+friction, if it's in the air.
Explanation:
Answer:
Explanation:
D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m
ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s
We can apply the equation
Ff = W ⇒ μ*N = m*g <em>(I)</em>
then we have
N = Fc = m*ac = m*(ω²*R)
Returning to the equation <em>I</em>
<em />
μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)
Finally
μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379
<span>Social
i think so ,but i am not sure</span>
Answer:
The work done is "2000 J".
Explanation:
The given values are:
Force,
F = 200 N
Mass,
m = 55 kg
Displacement,
d = 10 m
Now,
The work done will be:
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
