On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
1 answer:
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.
First of all, we need to calculate the speed of sound at temperature of
:
The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
First of all, we need to calculate the speed of sound at temperature of
The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
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magnetic flux ΦB = 0.450324 × weber
current I = 1.02484 A
Explanation:
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length a = 2.2 cm = 0.022 m
width b = 0.80 cm = 0.008 m
Resistance R = 0.40 ohms
current I = 4.7 A
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solution
we will find magnitude of magnetic flux thorough this formula that is
ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]
here μ is 4π × put all value
ΦB = (4π × 4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]
ΦB = 0.450324 × weber
and
current induced is
current = ε / R
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
put all value
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
current = 4π × (4.7) (0.022) (0.008) (0.0032) / 2π(0.40) [(0.012² ) - (0.008/2 )² ]
current = 1.02484 A