A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
Solution :
a). B at the center :
Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.
Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE
b). Also, the sum of the fields must be zero.
Therefore,
So,
A
Therefore, the current in the outer wire is 24.38 ampere.
You use more significant figures. 5 sigfigs (1.0985) is more accurate than 2 sigfigs (1.0)
The acceleration is 3.3 m/s2
Explanation:
Can be safer and cheaper than the real world. Able to test a product or system works before building it. Can use it to find unexpected problems. Can speed things up or slow them down to see changes over long or short periods of time.
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