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physixx help needed
- Thread starter Mueez
- Start date
Q 3....power on y-axis and time on x-axis
the area under the graph between two points, lets say at t=0 and t=10
so the area under the graph will be a triangle
the area of the triangle is 1/2 * (x-axis) * (y-axis)
so its time x power.......which gives u energy
so therefore its [A]
Q4...10ms---1cm
there are 8.5 waves in total in 6 cm
so 10----1cm
----x-----6
u cross multiply and get x as 60ms
60ms= 0.06 sec
0.06----8.5 waves
x--------1 wave
x= 0.06/8.5=0.00706 sec
u know f=1/T
so f=141 Hz...so its B
Q27...energy is force*distance
the distance moved by the positive charge is zero
so its A.
the area under the graph between two points, lets say at t=0 and t=10
so the area under the graph will be a triangle
the area of the triangle is 1/2 * (x-axis) * (y-axis)
so its time x power.......which gives u energy
so therefore its [A]
Q4...10ms---1cm
there are 8.5 waves in total in 6 cm
so 10----1cm
----x-----6
u cross multiply and get x as 60ms
60ms= 0.06 sec
0.06----8.5 waves
x--------1 wave
x= 0.06/8.5=0.00706 sec
u know f=1/T
so f=141 Hz...so its B
Q27...energy is force*distance
the distance moved by the positive charge is zero
so its A.
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oye mueez my battery got finished so cud not reply, yar we got resultant amplitude by subtracting the amplitudes of the given two waves and then calculated k from c i and used here to find the resultant intensity. did with the same method as libra94 used. . but hassam logic is also right .