An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. The car travels to the bottom, through a loop, an
d continues up the next hill. The end of the roller coaster has a level surface that is 4m off the ground. Assume there is no friction on the roller coaster ridea and energy is concerved. Q. How tall is the hill if the roller coaster is traveling at 10m/s at the top of it?
Q. What will be the velocity of the roller coaster at the top of the loop?
Let's start with the total amount of energy available for the whole scenario: Some kind of machine gave the coaster a bunch of potential energy by dragging it up to the top of a 45m hill,and that's the energy is has to work with.
So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.
KE = (1/2) (M) (V)² = 186,720
(400) (V)² = 186,720
V² = 186,720 / 400 = 466.8
V = √466.8 = <em>21.61 m/s</em> ===============================
Now it looks like there should be another question ... that's why they bothered to tell you that the end is 4m off the ground. They must want you to find the coaster's speed when it gets to the end.
At 4m off the ground, PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.
The rest will be kinetic. KE = (359,200 - 31,360) = 327,840 joules
KE = (1/2) (M) (V)² = 327,840
400 V² = 327,840
V² = 327,840 / 400 = 819.6
V = √819.6 = <em>28.63 m/s</em> at the end =======================================
If the official answers in class are a little bit different from these, it'll be because they used some different number for Gravity. I used '9.8' for gravity, but very often, they use '10' .
If the official answers in class are way way different from these, then I made one or more big mistakes somewhere. Sorry.
This is because distance traveled (i.e. displacement) is the integral of the velocity function, and velocity is the first derivative of the displacement function. To put this in perspective, the area bounded by a curve can be found by taking the integral of the equation of the curve, taking values on the x-axis as limits.
•To play Dr. Dodgeball you need to have 2 teams to verse each other. •Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors) •Continue to play dodgeball how you normally would •When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge. •Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends) Hope this helped! Play on! And plz mark brainliest lol this was long to write :D