A circular loop of wire with ra dius 8.0 cm carries a current of 0.20 A. A vec- tor of unit length and parallel to the dipole mo
- ment ii of the loop is 0.60i 0.80j. If the loop is located in a uniform magnetic field given by B (0.25T) (0.30T)k, what is the orientation energy of the loop? a) 0 J b) -6.0 x 10-4 J c) 2.5 x 10-4 J d) 3.2 x 10-4 J e) 5.1 x 10-3 J
1 answer:
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Answer:
The maximum speed will be 26.475 m/sec
Explanation:
We have given mass of the toy m = 0.50 kg
radius of the light string r = 1 m
Tension on the string T = 350 N
We have to find the maximum speed without breaking the string
For without breaking the string tension must be equal to the centripetal force
So
So
v = 26.475 m /sec
So the maximum speed will be 26.475 m/sec
Answer:0.13ml
Explanation:
