Answer:
66 rpm
Explanation:
The period of oscillation is given by
where T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.
Also, net force is given by
Net force=
where 
is the elongation, L is original length, 
is the angular velocity
Substituting the equation of 
into the above we obtain
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen
Neglect air resistance.
The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J
The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J
Answer: d. It has decreased by 200 J
Explanation:
The net force of each square is the combination of the forces in each direction. The direction is the... direction the square would go in due to the net force. The magnitude of the net force is how large it is. So if you had a force pushing 2N to the left and 1N to the right, then the net force would be 1N to the left; because the two oppose eachother. If they were going in the same direction, then they'd add to each other. And perpendicular net forces (like one pushing up and another pushing left) can create net forces in diagonal directions.
I'm not going to do all of these for you because they're basically all the same thing and it's good practice for you anyway. But I'll do the first three just so you can get the idea:
1. The net force's magnitude is 4N and it's direction is to the right.
2. The net force's magnitude is 4N and it's direction is to the left.
3. The net force's magnitude is 0N and it has no direction because they are equal forces acting in opposite directions.
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
