B is the correct answer for sure bro
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is
velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)
The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s.
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.
Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)
60 = 9.8t
60 / 9.8 = t
t = 6.122 s
Hopefully this is the right answer.
Answer:
D. I stayed up late for several nights. Eventually, I was too tired to concentrate.
Explanation:
i got it right edge 2020.
Explanation:
First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.
Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.
If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.
If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.
If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.
(see diagram below)