Good evening everyone, State the law of conservation of energy .Thank u
2 answers:
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For a simple harmonic motion energy is given with:
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.
Answer:
current in loops is 52.73 μA
Explanation:
given data
side of square a = b = 2.40 cm = 0.024 m
resistance R = 1.20×10^−2 Ω
edge of the loop c = 1.20 cm = 0.012 m
rate of current = 120 A/s
to find out
current in the loop
solution
we know current formula that is
current = voltage / resistance .................a
so current = 1/R × d∅/dt
and we know here that
flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b
so
d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c
so from equation a we get here current
current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt
current = ( 4π××0.024 / 2π(1.20×
) × ln (0.024 + 0.012/0.012) × 120
solve it and we get current that is
current = 4 ×× 1.09861 × 120
current = 52.73 × A
so here current in loops is 52.73 μA