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natali 33 [55]
3 years ago
15

If the loop is then converted into a rectangular loop measuring 2.1cm on its shortest side in 6.50ms, and the average emf induce

d across the loop is 14.7V during this time period, what is A.the strength of the B-field
Physics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

B =34.5T

Explanation:

Given E = 14.7V, L= 2.1cm = 0.021m, t = 6.50ms = 6.50×10-³s

A = L² = 0.021² = 4.41×10-⁴m²

ω = 2π/t = 2π/(6.50×10-³) = 966.64rad/s

E = ωBA

B = E/ωA = 14.7/(966.64×4.41×10-⁴)

B = 34.5T

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Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

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(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

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t is the time

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Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

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v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

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We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

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Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

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