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3 years ago

11

You drop a 30 g pebble down a well. You hear a splash 2.7 s later. Ignoring air resistance, how deep is the well? Assume g = 9.8

m/s2.
a)0.79 m
b)36 m
c)71 m
d)790 m

1 answer:

julia-pushkina [17]3 years ago

3 0

Here, you need to apply second equation of Kinematics
S = ut + 1/2 at²
s = at²/2 [ As initial velocity (u) = 0 ]

Now, a = 9.8 m/s²
t = 2.7 s

Substitute their values in the formula:
s = 9.8 × (2.7)² / 2
s = 35.721 m

After rounding-off to the nearest hundredth value, it would be 36 m

Finally, your answer would be option B.

Hope this helps!

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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

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The centripetal acceleration is defined as:

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Where v is the velocity and r is the radius

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$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

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Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

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$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

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$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

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$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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