Answer:
see below
Explanation:
First: Leave a couple inches of wire loose at one end and wrap most of the rest of the wire around iron u-shaped bar and make sure not to overlap the wires.
Second:Cut the wire (if needed) so that there is about a couple inches loose at the other end too.
Third: Now remove about an inch of the plastic coating from both ends of the wire and connect the one wire to one end of a battery and the other wire to the other end of the battery.
The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by 
, where 
here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving
You do 1000 divide it by 10 which equals 100 W
Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.
Answer:
The thrown rock will strike the ground 
earlier than the dropped rock.
Explanation:
<u>Known Data</u>
, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use 
, to find the total time of fall, so 
, then clearing for 
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use , to find the total time of fall, so 
, then, 
, as it is a second-grade polynomial, we find that its positive root is 
Finally, we can find how much earlier does the thrown rock strike the ground, so 
