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
Answer:
The angular speed of the object is 0.0281 rad/s
The linear speed of the object is 0.169 ft/s
Explanation:
Given;
radius of the circle, r = 6 ft
time of motion of the object around the circle, t = 80 s
central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad
The angular speed of the object is calculated as;
The linear speed of the object is calculated as;
v = ωr
v = 0.0281 rad/s x 6ft
v = 0.169 ft/s
Acceleration=force/mass=28/(10+4)=2m/s^2
force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block
vf=vi+at
-10=20*28/14 * t
t=30/2=15sec
i hope this can help you.
Answer:
Explanation:
Given that,
Frequency of a radio antenna is 1 MHz
Power, P = 21 kW
We need to find the the waves intensity 25 km from the antenna
. The object emits intenisty evenly in all direction. It can be given by :
So, the wave intensity 25 km from the antenna is 
.
Answer:
The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
Explanation:
The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.
Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.
From the law of conservation of energy, U₁ + K₁ = U₂ + K₂
So, kQ²/d + 0 = kQ²/3d + K
K₂ = kQ²/d - kQ²/3d = 2kQ²/3d
So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
