Number the steps for balancing equations: Use coefficients to increase the atoms on each side. Check to make sure you have the s
2 answers:
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Answer:
a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.
b) The projectile reaches maximum height 2.192 seconds after launch.
c) The maximum height of the projectile is 26.584 meters above ground.
d) The projectile will hit the ground at 4.523 seconds after launch.
e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.
Explanation:
Complete statement of problem is: <em>The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is </em><em>after t seconds. (Round your answers to two decimal places.) </em><em>(a)</em><em> Find the velocity after 2 seconds and after 4 seconds, </em><em>(b)</em><em> When does the projectile reach its maximum height? </em><em>(c)</em><em> What is the maximum height? </em><em>(d)</em><em> When does it hit the ground? </em><em>(e)</em><em> With what velocity does it hits the ground?</em>
a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:
(Eq. 1)
Where is the velocity function, measured in meters per second.
Now we evaluate this function at given times:
t = 2 s.
The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.
t = 4 s.
The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.
b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for :
The projectile reaches maximum height 2.192 seconds after launch.
c) Maximum height is calculated by evaluating height function at the time found in b). That is:
The maximum height of the projectile is 26.584 meters above ground.
d) In this case, we need to equalize the height function to zero and solve for . That is:
Roots are found by means of Quadratic Formula:
and
Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.
e) This can be found by evaluating velocity function at the time found in d):
The velocity of the projectile right before hitting the ground in -22.871 meters per second.
Answer: Choice B
There are triple bonds between the carbon (C) and oxygen (O) atoms. Then there are 2 dots on either side
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Explanation:
When it comes to interaction and chemistry, all that matters is the valence shell or valence electrons. This is the outermost shell. This is because various elements do not interact with the inner electrons (they're locked in place so to speak and don't move to other elements).
Carbon has 6 protons, which is what uniquely makes up this element. This means there are 6 electrons. The inner shell has 2 electrons and the valence shell has 4 electrons. Two electrons are shown as the two blue dots on the left side of the C. The other two electrons form two of the lines, or the bonds, between the C and O.
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Oxygen has 8 protons and 8 electrons. It has 2 electrons in the inner shell and 6 electrons in the valence shell. Two of those electrons are the red dots on the right side of the O. The other 4 electrons are shared to form the bonds with the carbon atom.
This is where things get a bit tricky. I've shown a diagram below indicating that one of the oxygen electrons (red dot) is passed to the carbon, as this carbon atom is pulling on the oxygen electron. But the oxygen atom is pulling on it as well, which forms one of the triple bonds.
So this is why diagram B is the final answer. This is something you can logically determine (remembering the rules of how each electron shell is formed), or it's something you'll need to memorize. In the real world, it's easy to forget a lot of info like this, so that's why having it handy as a lookup reference is preferable.
