A spring is compressed 8 cm. It is then brought back to equilibrium and compressed 24 cm. How much more work is done in the seco
1 answer:
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Answer:
32.46m/s
Explanation:
Hello,
To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are the follow
Where
Vf = final speed
Vo = Initial speed =7.3m/S
A = g=acceleration =9.81m/s^2
X = displacement =51m}
solving for Vf
the speed with the ball hits the ground is 32.46m/s
Answer:
757,93 feets
Explanation:
We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.
The angle of depression, its O in the graph.
Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.
ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).
Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.
Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.
We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:
This its equal to the distance from the boat to the shoreline.
