Answer: A) It has the same types of atoms on both sides of the reaction
arrow.
Explanation: A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.
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.
Answer:
![v = 4.7 m/s](https://tex.z-dn.net/?f=v%20%3D%204.7%20m%2Fs)
Explanation:
As we know that the ball is swinging in a circle
so here net horizontal centripetal force on the ball is due to component of the tension force
so we will have
![T sin\theta = m\omega^2 r](https://tex.z-dn.net/?f=T%20sin%5Ctheta%20%3D%20m%5Comega%5E2%20r)
![T cos\theta = mg](https://tex.z-dn.net/?f=T%20cos%5Ctheta%20%3D%20mg)
also we know that
![T = Mg](https://tex.z-dn.net/?f=T%20%3D%20Mg)
so we have
![Mg cos\theta = mg](https://tex.z-dn.net/?f=Mg%20cos%5Ctheta%20%3D%20mg)
![\theta = \frac{1}{2}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
![\theta = 60 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2060%20degree)
now we have
![Mg sin\theta = m\omega^2 (Lsin\theta)](https://tex.z-dn.net/?f=Mg%20sin%5Ctheta%20%3D%20m%5Comega%5E2%20%28Lsin%5Ctheta%29)
![Mg = m\omega^2 L](https://tex.z-dn.net/?f=Mg%20%3D%20m%5Comega%5E2%20L)
![2 \times 9.81 = 1 \omega^2 (1.5)](https://tex.z-dn.net/?f=2%20%5Ctimes%209.81%20%3D%201%20%5Comega%5E2%20%281.5%29)
![\omega = 3.62 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%203.62%20rad%2Fs)
now we have
![v = r\omega](https://tex.z-dn.net/?f=v%20%3D%20r%5Comega)
![v = (1.5 sin60) (3.62)](https://tex.z-dn.net/?f=v%20%3D%20%281.5%20sin60%29%20%283.62%29)
![v = 4.7 m/s](https://tex.z-dn.net/?f=v%20%3D%204.7%20m%2Fs)
The speed should be slower at the top of the ramp and gain speed toward the bottom of the ramp