Answer:
The electric potential is 
Explanation:
From the question we are told that
The length of the rod is 
The total charge of the rod is 
The length from the center is 
The diagram illustrating the setup for this question is shown on the first uploaded image
From the diagram the potential at point A 
is mathematically represented as
Where K is the coulomb constant with a value 
where q is the charge in charge the rod relative to its distance from A is mathematically represented as
This a small unit length of the rod
So 
=> ![V = k\frac{q}{L} ln [\frac{4}{2} ]](https://tex.z-dn.net/?f=V%20%3D%20%20k%5Cfrac%7Bq%7D%7BL%7D%20%20ln%20%5B%5Cfrac%7B4%7D%7B2%7D%20%5D)
Substituting values
Answer:
Explanation:
As we know that the mass is revolving with constant angular speed in the circle of radius R
So we will have
now the position vector at a given time is
now the linear velocity is given as
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
<h2>
Answer:</h2>
-310J
<h2>
Explanation:</h2>
The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e
ΔE = Q + W ----------------------(i)
If heat is released by the system, Q is negative. Else it is positive.
If work is done on the system, W is positive. Else it is negative.
<em>In this case, the system is the balloon and;</em>
Q = -0.659kJ = -695J [Q is negative because heat is removed from the system(balloon)]
W = +385J [W is positive because work is done on the system (balloon)]
<em>Substitute these values into equation (i) as follows;</em>
ΔE = -695 + 385
ΔE = -310J
Therefore, the change in internal energy is -310J
<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>
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No because the whether changes constant, weather can be close but is not the 100% the correct answer,..
