Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>
![y_{f}=0m](https://tex.z-dn.net/?f=y_%7Bf%7D%3D0m)
![v_{iD}=0m/s](https://tex.z-dn.net/?f=v_%7BiD%7D%3D0m%2Fs)
, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so ![t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s](https://tex.z-dn.net/?f=t_%7BE%7D%3Dt_%7BD%7D-t_%7BT%7D%3D7.82s-5.4s%3D2.42s)
Given,
Weight of a block on the Earth = 980 N
Acceleration due to gravity on Earth = 9.8 N/kg
To find,
Weight of the block on the moon.
Solution,
The moon has acceleration (1/6) times that on earth. Let m be the mass of the block on the moon.
We know that,
W = mg
![m=\dfrac{W}{g}\\\\m=\dfrac{980}{9.8}\\\\=100\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BW%7D%7Bg%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B980%7D%7B9.8%7D%5C%5C%5C%5C%3D100%5C%20kg)
As mass of an object remains the same everywhere. Weight of the block on Moon is :
W' = mg'
As g'=(g/6)
So,
![W=100\times \dfrac{g}{6}\\\\=100\times \dfrac{9.8}{6}\\\\=163.33\ N](https://tex.z-dn.net/?f=W%3D100%5Ctimes%20%5Cdfrac%7Bg%7D%7B6%7D%5C%5C%5C%5C%3D100%5Ctimes%20%5Cdfrac%7B9.8%7D%7B6%7D%5C%5C%5C%5C%3D163.33%5C%20N)
So, the weight of the block on the moon is 163.33 N.
Answer:
to the left.
Explanation:
We can use the equation
where v is the velocity at time t and
the velocity at
. Since we want the acceleration we write this equation as:
![a=\frac{v-v_0}{t-t_0}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-v_0%7D%7Bt-t_0%7D)
Considering the <em>direction to the right as the positive one</em>, we have
at
, and
at
, so we substitute:
![a=\frac{v-v_0}{t-t_0}=\frac{(-8m/s)-(5m/s)}{(10s)-(8s)}=\frac{-13m/s}{2s}=-6.5m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-v_0%7D%7Bt-t_0%7D%3D%5Cfrac%7B%28-8m%2Fs%29-%285m%2Fs%29%7D%7B%2810s%29-%288s%29%7D%3D%5Cfrac%7B-13m%2Fs%7D%7B2s%7D%3D-6.5m%2Fs%5E2)
Where the minus sign indicates it is directed to the left.
Answer: As Earth rotates, the Moon's gravity causes the oceans to seem to rise and fall. ... There is a little bit of friction between the tides and the turning Earth, causing the rotation to slow down just a little. As Earth slows, it lets the Moon creep away.
Explanation:
The answer would be aircraft carrier