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3 years ago

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A grandfather clock keeps time using a pendulum. Which will be true of the clock if the pendulum is shortened? Question 10 optio

ns:
It will run slow.
It will run fast.
It will keep the same time.
It will depend on the bob’s mass.

1 answer:

katovenus [111]3 years ago

4 0

Answer: The period of the pendulum is shorter if the pendulum itself is shortened.

Explanation: i did "It will depend on the bob’s mass". and got it wrong so it's not that one haha

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A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

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2 years ago

A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a

Schach [20]

Before it hits the sand bed, the meteorite is accelerating uniformly with $g=9.80\dfrac{\rm m}{\mathrm s^2}$ , so that its speed $v$ satisfies

$v^2-{v_0}^2=-2g\Delta y$

where $v_0=90.0\dfrac{\rm m}{\rm s}$ is its initial speed and $\Delta y=(0-850)\,\mathrm{km}=-850\,\mathrm{km}$ is its change in altitude. Notice that we're taking the meteorite's starting position in the atmosphere to be the origin, and the downward direction to be negative. Now,

$v^2-\left(-90.0\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(-850,000\,\mathrm m)\implies\boxed{v=4080\dfrac{\rm m}{\rm s}}$

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3 years ago

What forces cause the Sun’s magnetic field to become both stronger and more tangled

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The gravitational pull causes the sun’s magnetic field to become both stronger and more tangled

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3 years ago

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma

tresset_1 [31]

Answer:

Explanation:

a. Given that:

$m-$ mass of first & second piece, $3m-$ mass of 3rd piece, $\bar v_1$ -velocity of first piece( $-23\dot i \ m/s$ ) and $\bar v_2$ as velocity of 2nd piece ( $-23\dot j \ m/s$ )

Let $\bar v_3$ be velocity of 3rd piece=?

#Vessel is at rest before explosion. Considering conservation of linear momentum:

$m\bar v_1 +m\bar v_2 +3m \bar v_3=0$ #Dividing both sides by $m, m\neq 0$

$\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)$

#Plug the $\bar v_1 ,\bar v_2$ values:

$\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)$

#So the magnitude of the third piece is:

$|\bar v_3|=\sqrt{(23/3)^2+(23/3)^2}\\=10.84m/s$

Magnitude of the 3rd piece is 10.84 m/s

b. To find direction of the magnitude (as an angle relative to the $x$ -axis), we find $\angle \theta$ . The angle is obtained by getting the tan inverse as:

$\theta=tan^-^1(\frac{23/3}{23/3})\\=45\textdegree$

-The direction of the magnitude (angle relative to the x-axis) is 45°

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3 years ago