One of two things is true about this question: EITHER it can't happen
as you've described it, OR you've left out some vital information.
-- IF the first stone was thrown downward with an initial speed and the
second one was dropped from rest 1 second later, then the second one
can never catch up with the first one, and they can never hit the water together.
-- IF the first stone was thrown downward with an initial speed, AND the
second one was released 1 second later, AND they actually do hit the
water together, THEN the second stone must have been given an initial
downward speed greater than 2 m/s, otherwise it could never catch up
with the first one.
Note:
The masses and weights of the stones are irrelevant and not needed.
=======================================================
An afterthought . . . . .
If the first stone was tossed UP at 2 m/s . . . that could be the meaning of the
prominent plus-sign that you wrote next to the 2 . . . then it rises for (2/9.8) second, then begins to fall, and passes the mountain climber's hand on the way down (4/9.8) second after he tossed it, falling at the same 2.0 m/s downward.
From there, it still has 50m to go before it hits the water.
50 = 2 T + 1/2 G T²
4.9 T² + 2 T - 50 = 0
T = 3 seconds
The first stone hits the water 3 seconds after passing the mountain climber's hand on the way down at a downward speed of 2.0 m/s. In that 3 seconds, it gains (3 x 9.8) = 29.4 m/s of additional speed, hitting the water at (29.4 + 2) = 31.4 m/s .
This is all just a guess, assuming that the 2.0 m/s was an UPWARD launch.
Maybe I'll come back later and calculate the second stone.
A geologist is studying rock layers in an old river bed, and he finds a fossil of a fish and a horsetail rush in the same rock layer. According to the law of faunal and floral succession, the geologist can assume that the rock containing the fossils may date back as far as the <span>Devonian period</span>.
Answer:
A measurement standard is a quantity that people agree to use as a comparison. Standards are important because they allow measurements to be compared even if different people in different parts of the world take them.
Hope this helps ⊂◉‿◉つ
Answer:
Velocity will be ![v=4.06\times10^7m/sec](https://tex.z-dn.net/?f=v%3D4.06%5Ctimes10%5E7m%2Fsec)
Explanation:
We have given mass of alpha particle m = 4 u ![=4\times 1.67\times 10^{-27}kg=6.68\times 10^{-27}kg](https://tex.z-dn.net/?f=%3D4%5Ctimes%201.67%5Ctimes%2010%5E%7B-27%7Dkg%3D6.68%5Ctimes%2010%5E%7B-27%7Dkg)
Charge on alpha particle ![q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C](https://tex.z-dn.net/?f=q%3D2e%3D2%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7DC%3D3.2%5Ctimes%2010%5E%7B-19%7DC)
Charge on thorium particle ![=90\times 1.6\times 10^{-19}=144\times 10^{-19}C](https://tex.z-dn.net/?f=%3D90%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%3D144%5Ctimes%2010%5E%7B-19%7DC)
Diameter is given as d = 15 fm
So radius ![r=\frac{d}{2}=\frac{15}{2}=7.5fm=7.5\times 10^{-15}m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B15%7D%7B2%7D%3D7.5fm%3D7.5%5Ctimes%2010%5E%7B-15%7Dm)
Potential energy is given by ![E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r}=\frac{Kq_1q_2}{r}=\frac{9\times 10^9\times 144\times 10^{-19}\times 3.2\times 10^{-19}}{7.5\times 10^{-15}}=5.529\times 10^{-12}J](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%20_0%7D%5Cfrac%7Bq_1q_2%7D%7Br%7D%3D%5Cfrac%7BKq_1q_2%7D%7Br%7D%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20144%5Ctimes%2010%5E%7B-19%7D%5Ctimes%203.2%5Ctimes%2010%5E%7B-19%7D%7D%7B7.5%5Ctimes%2010%5E%7B-15%7D%7D%3D5.529%5Ctimes%2010%5E%7B-12%7DJ)
From energy conservation ![\frac{1}{2}mv^2=5.529\times 10^{-12}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D5.529%5Ctimes%2010%5E%7B-12%7D)
![\frac{1}{2}\times 6.68\times 10^{-27}v^2=5.529\times 10^{-12}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%206.68%5Ctimes%2010%5E%7B-27%7Dv%5E2%3D5.529%5Ctimes%2010%5E%7B-12%7D)
![v=4.06\times10^7m/sec](https://tex.z-dn.net/?f=v%3D4.06%5Ctimes10%5E7m%2Fsec)