The induced emf is proportional to the rate of change of flux (tangent of the graph). since there is a change of flux and the rate is a constant (straight line curve ), the induced emf is of a constant value.. thus the horizontal line.
The induced emf is proportional to the rate of change of flux (tangent of the graph). since there is a change of flux and the rate is a constant (straight line curve ), the induced emf is of a constant value.. thus the horizontal line.
For question 13) What would i use to solve it? For question 3) Why isnt it the area under the graph, which gives change of momentum?
13. Grav potential difference is work done per kg. If it's 3 J/kg over 10m and the field is uniform it's 1.5 J/kg over 5m. You are given the mass in kg so what is the number of joules needed over that 5m?
13. Grav potential difference is work done per kg. If it's 3 J/kg over 10m and the field is uniform it's 1.5 J/kg over 5m. You are given the mass in kg so what is the number of joules needed over that 5m?
3. Is the given answer not C?
Ahh yes it is thanks! I got confused with another question.
As a poster in the thread has said, Ln V against t gives a straight line. The maths tells you this. Take logs of both sides of the formula for discharge. You get y = mx +c with negative m.
As a poster in the thread has said, Ln V against t gives a straight line. The maths tells you this. Take logs of both sides of the formula for discharge. You get y = mx +c with negative m.