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3 years ago

A circular wire loop of radius 15.0 cm carries a current of 2.60

1 answer:

5 0

Part (a): Magnetic dipole moment

Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2

Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm

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A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T

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Let $y_0$ be the height of the building and thus the initial height of the ball. The ball's altitude at time $t$ is given by

$y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the ground when $y=0$ after $t=2.9\,\mathrm s$ . Solve for $y_0$ :

$0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2$

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so the building is about 16 m tall (keeping track of significant digits).

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2 years ago

(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the

frosja888 [35]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a. $C=8\mu F=8\times 10^{-6} F$

$1\mu =10^{-6}$

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We know that

$V=\frac{Q}{C}$

Using the formula

$V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V$

b. $Q=192\mu C=192\times 10^{-6} C$

$V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V$

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