A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away at a rate of 2 ft/sec.
a.) How fast is the top of the ladder moving down the wall when the base of the ladder is 12 feet from the wall?
Answer:
dy/dt = -1.094ft/sec
Explanation:
Given that:
dz/dt = 0,
dx/dt = 2,
dy/dt = ?
Hence, we have the following
Using Pythagoras theorem
We have 25ft as the hypotenuse, y as the opposite or height of wall, and x as the base of the triangle
X² + y² = z²,
12² + y² = 25²,
y² = 25² - 12²
y = √481
Therefore, we have the following:
2x dx/dt + 2y dy/dt,
= 2z dz/dt,
= 12 (2) √481 dy/dt,
= √481 dy/dt = -24,
= dy/dt = -1.094ft/sec
Therefore, final answer is -1.094ft/sec
Here we go.
My abbreviations; KE = Kinetic Energy; GPE = Gravitational Potential Energy.
So first off, we know the fish has KE right when the bird releases it. Why? Because it has horizontal velocity after released! So let’s calculate it:
KE = 1/2(m)(V)^2
KE = 1/2(2)(18)^2
KE = 324 J
Nice!
We also know that the fish has GPE at its maximum height before release:
GPE = mgh
GPE = (2)(9.81)(5.40)
GPE = 105.95 J
Now, based on the *queue dramatic voice* LAW OF CONSERVATION OF ENERGY, we know all of the initial energy of the fish will be equal to the amount of final energy. And since the only form of energy when it hits the water is KE, we can write:
KEi + GPEi = KEf
(Remember - we found the initial energies before!)
(324) + (105.95) = KEf
KEf = 429.95J
And that’s you’re final answer! Notice how this value is MORE than the initial KE from before (324 J) - this is because all of the initial GPE from before was transformed into more KE as the fish fell (h decreased) and sped up (V increased).
If this helped please like it and comment!
Answer:
An atom that loses a electron is called a cation and has an overall positive charge.
