Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s
Answer:
sorry I don't really know :P
Explanation:
The answer is: [C]: Neither Juan nor Christina are correct.
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Explanation:
_____________
Consider Juan's statement:
_______________________
"Juan said fossils deposits always contain only one type of organism."
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This statement is incorrect. There are many instances of fossil deposits containing more than one type of organism.
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Consider Christina's statement.
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"Christina said fossil deposits never contain one type of organism."
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This statement is incorrect. Fossil deposits, by definition (of "fossil") ALWAYS contain <u>at least one type</u> of organism.
___________________________________________
As such, neither Juan nor Christina are correct.
And as such, Answer choice: [C] is the correct answer.
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Answer choices A, B, and D are ruled out.
Answer:
The diameter of the needle is <u>4.675 cm</u>.
Explanation:
Given:
Volume flow rate is, 
Velocity of air expelled by pump is, 
Let the area of the needle be 'A' cm² and the diameter be 'd' cm.
We know that, volume flow rate of the air expelled by pump is given as the product of the needle's area and velocity of air flowing through that area.
Therefore, volume flow rate is given as:
Now, considering the needle to be circular, area of the needle can be written as:
Therefore, the diameter of the needle is 4.675 cm.
