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>
When ignited, the gas mixture converts to water vapor and releases energy, which sustains the reaction: 241.8 kJ of energy (LHV) for every mole of H2 burned.” A mole of hydrogen weighs 2 grams. So, this is a LHV (lower heating value) of 120.9 kJ/gram of hydrogen when heat of vaporization is subtracted.
Answer:
Explanation:
Assuming the ground is level as well.
F = ma
a = F/m
a = (2000 - 350) / 1500
a = 1.1 m/s²
