Answer:
b
Explanation:
because if you freeze somthing you do not gain heat you loss heat.
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>
Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:
By using the transformer formula, we get
⇒ 
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Answer:in X-rays
Explanation:
Radiation are used in X-rays for the treatment of tumours, and some skin diseases,cancer cells can be destroyed by X-rays
