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Soloha48 [4]
3 years ago
10

What is the value of sin square 60​

Physics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

sin {}^{2} 60 =  \frac{3}{4}

Explanation:

sin^2 60° = ( \|3 / 2 ) ^2 = 3 / 4.

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Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea
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What is the total current from the battery? *<br> {R,<br> {R<br> &amp;R,<br> 612<br> 7<br> 6<br> 5
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For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo
Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}

so that its period is T=3.644\,\frac{\mathrm s}{\mathrm{rev}} (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration \Delta t. Denote by \theta the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ

\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ

\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ

We can then compute the magnitude of the velocity vector differences \Delta\vec v for each time interval by using the law of cosines:

|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta

\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}

and in turn we find the magnitude of the average acceleration vectors to be

\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0
3 years ago
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