Physics Oscilloscope Question

It's a power question. You know R is 100 and in the first 1ms that V is -5, so you can use P=(V^2)/R
Do the same for the other time divisions

It's not a power question. It's asking for the amount of energy dissipated.

Energy = Power x time

Ty, forgot to add this part on. You can work out the power using V and R and then multiply by t to get the energy

Total power dissipated in 1 cycle = Power dissipated in 1 cycle

(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle

Yeah but how do you find the power dissipated in 1 cycle. Do you add up the power (in which case power would equal 0)? Do you multiply power by time? Do you find an average for power by dividing power by the number of horizontal lines? What do you do?

Hi, can anyone help me with part b of this question? I'm not sure what equations I'm supposed to use.

https://imgur.com/a/7jo2BNv

Total power dissipated in 1 cycle = Power dissipated in 1 cycle
(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle

becomes 25.0 V^2 and

Mean power is just mean power. In kinematics, we can say average velocity which is found by dividing the total displacement by total time taken. Based on the above writing,


(IMO) It is a confusing way of writing.

They just asked how to find to find answer to (b)(iv) in post 9. I'm very sorry you find it confusing. It is the mean power dissipated = total power dissipated in 1 cycle / time for 1 cycle.

I don't understand what is meant by "velocity occur in 1 cycle".

Look at your writing, it originates from you.
(Mean) power dissipated in 1 second = Power dissipated in 1 cycle / Time for 1 cycle
(Average) velocity in 1 second = velocity occur in 1 cycle / Time for 1 cycle

They just asked how to find to find answer to (b)(iv) in post 9. I'm very sorry you find it confusing. It is the mean power dissipated = total power dissipated in 1 cycle / time for 1 cycle.

Ok

If you read OP’s questions in post #9 again, the questions are raised in regard to your previous post #8. Don’t really know what you are trying to say in post #8. I believe in A-level physics, you have learned

and this should be used to find mean power

Power will not be equal to 0.

P = V^2/R = I^2 R which are never negative.

Mean power = Energy in 1 cycle / Time for 1 cycle
Use the time base and y-gain.

Energy dissipated in 1 cycle = Energy during 0 to 1 ms + Energy during 1 to 1.5 ms because first 1.5ms is one cycle

so ans to (b)(iii) = ans to (b)(i) + ans to (b)(ii)

Edited to explain mean power = energy dissipated in 1 cycle / time for 1 cycle:

Power dissipated in 1 cycle = energy dissipated in time for 1 cycle / time for 1 cycle = energy dissipated in T seconds / T

Mean power dissipation = energy dissipated in T seconds / Time for 1 cycle = energy dissipated in T seconds / T

ans to (b)(iv) = energy dissipated in 1 cycle / time for 1 cycle = ans to (b)(iii) / time for 1 cycle = ans to (b)(iii) / period of signal = ans to (b)(iii) x frequency of signal = ans to (b)(iii) x ans to (a)(iii)



Average velocity, = $\displaystyle \dfrac{1}{t_2-t_1} \int_{t_1}^{t_2} v(x) dx$
e.g $v(t) = 2$

Then average velocity for t = 3 to t = 5, $v = \displaystyle \dfrac{1}{5-3} \int_{3}^{5} 2 dx = 2$

If motion is periodic and it is meaningful to talk of cycles, then v = displacement over 1 cycle / time for 1 cycle

I do not understand "velocity occur for 1 cycle".

Can someone just tell me how to find energy dissipated in one cycle and energy dissipated in one second. It's only one / two marks per part so I shouldn't need to integrate anything.

See post #11
https://www.thestudentroom.co.uk/showpost.php?p=81258348&postcount=11