Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
B, do earthworms prefer bright light or darkness!
(a) 392 N/m
Hook's law states that:
(1)
where
F is the force exerted on the spring
k is the spring constant
is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:
- The stretching of the spring is
Solving eq.(1) for k, we find the spring constant:
(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is
And so, the stretch of the spring is
And since the initial lenght of the spring is
The final length will be
Answer:x(t)= Acos(wt)
Explanation:
According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx
Where x is displacement from equilibrium
K = spring constant
Therefore X(t) = Acos(2pit/T)
X(t)= Acos(wt)
Well it's energy comes from the sun since the sun gives it sunlight to help it grow
