Ok i apologise for the messy working but I'll try and explain my attempt at logic
Also note i ignore any air resistance for this.
First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.
Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.
I also noted the "initial" and "final" height of the swing with hi and hf respectively.
So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.
So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).
Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.
The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).
I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx
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Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²
= (1/2) (9.8 m/s²) (95 sec)²
= (4.9 m/s²) (9,025 sec²)
= 44,222.5 meters .
The depth of the mine shaft is five times the height of Mt. Everest !
Answer:
(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Explanation:
Given that,
Initial spinning = 50.0 rad/s
Time = 20.0
Distance = 2.5 m
Mass of pole = 4 kg
Angle = 60°
We need to calculate the angular acceleration
Using formula of angular velocity




The angular acceleration is -2.5 rad/s²
We need to calculate the number of revolution
Using angular equation of motion

Put the value into the formula


The number of revolution is 500 rad.
(II). We need to calculate the torque
Using formula of torque


Put the value into the formula


Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Answer:
D
Explanation:
Work is not a vector but it is a scalar