In Fig. 28-57 , the two ends of a U-shaped wire of mass m=...

In Fig. $28-57$ , the two ends of a U-shaped wire of mass $m=$ 10.0 $\mathrm{g}$ and length $L=20.0 \mathrm{cm}$ are immersed in mercury (which is a conductor). The wire is in a uniform field of magnitude $B=0.100 \mathrm{T}$ . A switch (unshown) is rapidly closed and then reopened, sending a pulse of current through the wire, which causes the wire to jump up- ward. If jump height $h=3.00 \mathrm{m}$ , how much charge was in the pulse? Assume that the duration of the pulse is much less than the time of flight. Consider the definition of impulse (Eq. $9-30$ ) and its relationship with momentum (Eq. $9-31 )$ . Also consider the relationship between charge and current (Eq. $26-2 )$ .

Transcript

00:01 In this question it is told that a landscape architect is planning for an artificial waterfall and we have given that water flowing at a 1 .7 meter per second and it leaves the end of a horizontal channel at the top of a vertical wall that is 35.

00:15 Sorry, this is 2 .35 meter high.

00:19 Now in the first part of the problem we have to find that will the space behind the waterfall be wide enough for a bedstrain walk away.

00:29 So first of all we can say that i'm going to draw the diagram for this cushion so this is the height of the wall and from here the water falls with a velocity that is given to us and this is the range of this horizontal range and this velocity next direction is given to us that is we not is equal to 1 .7 meter per second and this height is given to us that is 2 .35 meter and this is x now first of all we have to find this x.

01:11 So we can say that for this question we have given that there is an acceleration in vertical direction that is a y is equal to minus of 9 .8 meter per second is squared that is downward direction and acceleration is horizontal direction is zero.

01:25 Now first i am using equation in vertical motion that is yf can be written as y initial plus that is initial velocity in y direction into t plus of half acceleration in y direction into t from here we get the value of time taken by the water to fall.

01:44 So this is 0 is 2 .35 plus initial in wide direction there is no velocity so this is 0 plus half into this is minus 9 .8 into t square.

01:58 From here we get the value of t and this will come out to be 0 .6925 second...

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