Answer:
v = 7.32 m/s
Explanation:
The potential energy will convert to kinetic energy
½Iω² + ½mv² = mgh
Iω² + mv² = 2mgh
(½mR²)(v/R)² + mv² = 2mgh
½mv² + mv² = 2mgh
½v² + v² = 2gh
3v²/2 = 2gh
v² = 4gh/3
v² = 4(9.81)(4.10)/3
v² = 53.628
v = 7.323114...
v = 7.32 m/s
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "<span>The charge can be located anywhere, since flux does not depend on the position of the charge as long as it is inside the sphere" gives the answer to the question. I hope that this answer has come to your help.</span>
The rate of change of d(t) at t = 2 and t = 6 is the ratio between the change of distance (difference between the distances) to the time elapsed. That is,
r = (576 - 64) / (6 - 2) = 128 ft /s
The rate of change is equal to 128 ft/s and this represents the average speed at this time interval.
If the runner is running in a circular track then yes when something or someone is moving in a circular motion at a constant speed they are indeed accelerating. They’re accelerating because the direction of the velocity vector is changing
So first Identify all the given Varibales so u can choose which Eqauton to use
D=200m
T=4s
Vi=10m/s
Vf=?
You should this equation
D= 0.50(Vf+Vi)T
Plug in the values
200= 0.50 (Vf+10) 4
Divide the 4 out of the right side and if you do sumthing to one side you gotta do it to the other
200 divided by 4= 0.50(Vf+10)
50= 0.50(Vf+10)
Now expand the 0.50
So 50= 0.5Vf + 5 (because 0.5 times 10 is 5)
Now get rid of the 5
50-5= 0.5Vf
45 =0.5Vf now Divide the 0.5 out
45 divided by 0.5 = Vf
And 45/0.5 is 90
So 90=Vf
Therefore the final Velocity is 90m/s
