Answer:
Option (e) = 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.
Explanation:
So, we are given the following set of infomation in the question given above;
=> "spherical Gaussian surface of radius R centered at the origin."
=> " A charge Q is placed inside the sphere."
So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?
The CORRECT option (e) that is " 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." Is correct because of the reason given below;
REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.
Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.
Answer:
Explanation:
An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.
Answer:
Look to the explanation
Explanation:
<u><em>Work done</em></u> is is the energy transferred to or from an object by means
of a force acting on the object.
Work is positive if energy transferred to the object, and work is
negative if energy transferred from the object.
<em>Work = Force in the direction motion of object × its displacement</em>
The SI unit of the work is joule (J)
<u><em>Power</em></u> is the rate of work
<em>Power = work done ÷ time taken</em>
Power =
Displacement (s) ÷ time (t) = velocity (v)
<em>Power = Force × velocity</em>
The SI unit of the power is watt (w)
a)
We use the formula :
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the values in:
4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f
Calculating this we get:
32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f
Rearrange for v2f:
v2f =
This gives us 8.0 m/s as the final velocity of the second ball.
b)
Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.
This means we use the formula:
Ek = + = +
Substituting in values:
Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2
This simplifies to:
Ek= 128J + 0J = 0J + 128J
This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.
Answer:
a) the distances are zero, Both 1st & 2nd condition
c) the torques are equal but of the opposite sign, 2nd condition of equilibrium
Explanation:
The equilibrium conditions are
1 translational
∑ F = 0
2 rotational
∑ τ = Σ (F_i x r_i) = 0
They tell us that external torque is zero.
Therefore we have two various possibilities
a) the distances are zero, in this case we have a pure translation movement
for this situation the two equilibrium relations are fulfilled
b) the forces are zero, there is no movement
It does not make sense to use the equilibrium relations since there are no forces
c) the torques are equal but of the opposite sign, the forces are on the opposite side of the body.
In this case the 2 equilibrium relation is fulfilled, but not the first one that the force has the same direction