A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters
Answer:
the SI unit of momentum is :- kg.ms-1
and we know that,
kinetic energy = 1/2 mv2
E=p2/2m
p=(2Em)1/2
so the derived units are (J.kg)1/2
Explanation:
The question just basically explained what happens
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
Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N