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Basile [38]
3 years ago
11

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. Ass

ume the clock hand has a length of 15.0 cm and a mass of 6.00 g. Consider the second hand to be a slender rod rotating with constant angular velocity about one end. (Express your answer in the form: "1.23e-4 kg*m^2/s'.)
Physics
2 answers:
Lera25 [3.4K]3 years ago
4 0

Answer:

L=4.711×10^{-6}Kgm^2/sec

Explanation:

we have given mass m=6 gram =0.006 kg

length =15 cm =0.15 m

moment of inertia for the clock hand is given by

I=\frac{ml^2}{3}

I=\frac{0.006\times .15^2}{3}=4.5\times 10^{-5}

angular velocity \omega =\frac{2\times \pi}{60}=0.1047 rad/sec

we know that angular momentum is given by

L=Iω

L=4.5×10^{-5}×0.1047

L=4.711×10^{-6}Kgm^2/sec

Naddika [18.5K]3 years ago
3 0

Answer:

L = 4.711 *10^{-6} kg m2/s

Explanation:

i =\frac{ml^3}{3}

= \frac{0.00*.15^2}{3}

   =4.5*10^-5

angular velocity

\omega = \frac{2\pi}{60}

             = 0.1047 rad/s

the angular momentum,

L = I\omega

L = 4.5*10^{-5}* 0.1047 rad/s

L = 4.711 *10^{-6} kg m2/s

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Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

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F_d = mg

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' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

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= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

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