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>
The masses are equal.
According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
I hope that this helps you!
Answer:
p = 22.5 cm
Explanation:
For this exercise we must use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm
since the image is on the same side as the object according to the sign convention the distance is negative
f = -45.0 cm
now let's use the near point (q = 15.0 cm) at what distance the object should be
1 / p = 1 / -45 - 1 / -15
1 / p = -1/45 + 1/15
= 0.0444
p = 22.5 cm
this is the closest distance you can see an object clearly
<h2><u>Projectile</u><u> </u><u>motion</u><u>:</u></h2>
<em>If</em><em> </em><em>an</em><em> </em><em>object is given an initial velocity</em><em> </em><em>in any direction and then allowed</em><em> </em><em>to travel freely under gravity</em><em>, </em><em>it</em><em> </em><em>is</em><em> </em><em>called a projectile motion</em><em>. </em>
It is basically 3 types.
- horizontally projectile motion
- oblique projectile motion
- included plane projectile motion
