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3 years ago

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A woman is sitting on a spinning seat of a piano stool with her arms folded. Ignore any friction in the spinning stool. What hap

pens to her angular velocity and angular momentum when she extends her arms outward?

1 answer:

coldgirl [10]3 years ago

7 0

Answer:

The angular velocity of the woman will decrease when she extends her arms outwards.

Explanation:

We know that the angular momentum of a body is the product of its angular velocity and the moment of inertia.

So, mathematically:

$L=I.\omega$

where:

$I=$ moment of inertia (second moment of mass that depends upon the radial distance of the mass from the center of rotation)

$\omega=$ angular velocity

When the woman extends her arms she increases the radial distance of her mass form the axis of rotation thus increasing the moment of inertia of her body. As the angular momentum in this case remains constant so proportionately the angular velocity of her body increases.

$I'\times \omega'=I\times \omega$

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Answer:

The magnitude of the field is 8.384×10^-4 T.

Explanation:

Now, i start solving this question:

First, convert the potential difference(V) 2 kv to 2000 v.

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r = 0.36/2 = 0.18 m.

As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.

We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.

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These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:

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qB = mv/r

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A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −

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Answer:

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Explanation:

Given that,

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The electron's speed has fallen by half when it reaches x = 0.190 m.

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Change in potential energy $\Delta P.E=-9.60\times10^{-17}\ J$ as it goes x = 0.190 m to x = -0.210 m

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Using formula of work done

$W=-eE\Delta x$

Put the value into the formula

$W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))$

$W=-5.04\times10^{-17}\ J$

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Using change in kinetic energy,

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$\Delta K.E=\dfrac{-3mv^2}{8}$

Now, using work energy theorem

$\Delta K.E=W$

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Put the value in the equation

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$v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}$

Put the value of m

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Using change in energy

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$\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}$

Put the value into the formula

$v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}$

$v_{f}=1.5\times10^{7}\ m/s$

Hence, The speed of electron is $1.5\times10^{7}\ m/s$

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