Oh, so the time is 5s then. 2s after the explosion and 3s to come back down? If the total time is 5s then horizontal distance is 15m/s*5s = 75m and the total distance between the two fragments after they hit the ground is 33m+75m = 108m. That's the answer listed in the book.
Ok, so i figured out what I was doing wrong. The fragment moving left travels a vertical distance of 25m. Right before it starts coming back down, its velocity at the highest point would be 0. I used this info to calculate time on the downward flight:
Y = Vi*t + 1/2gt^2
25 = 0 +...
A firecracker with mass 100g is propelled vertically with a launch force of 1.2N applied over 5s, after which it explodes into two 50g fragments. One fragment goes horizontally at 15m/s and the other goes to the right at angle 53 degrees with the horizontal at speed 25 m/s. find the distance...
A 0.8 kg ball is whirled on a string r = 0.4 meters long in a vertical circular path. At the bottom of the circle, the ball is h = 0.45 meters from the ground. At the top of the circle, the ball has a speed of 3 m/s. Assume that the total energy of the ball is kept constant.
avg velocity comes to mind since the question says the elevator travels a distance x over a period of 10s. If i solve the problem using this information, the answer comes out to be 2X + 1000.
But avg accleration is equal to delta v/ delta t, not avg velocity. And also, using avg velocity would...
100 kg elevator acclerates up at rate avg a*. What's the avg force acting on the elevator if it covers a distance X over a period of 10s?
a* denotes average accleration
a. 2X + 1000
b. 100(a* + 1)
c. 2a* + 1000
I set up a free body diagram to show all the forces acting on the...
So confused about standing waves in a closed pipe, which is open at one end and closed at the other. The closed end has a node while the open end has an antinode. To figure the wavelength, i use the formula:
Lambda = 4L/n where n is the number of harmonic and can only be odd integers...
I used deltas to define the initial and final energies of the entire syste.
Ei = Ui + KEi
= 1/2kx^2 + 0
So, the initial energy = 1/2kx^2 only b/c the block has not been released yet.
Ef = (Uf + KEf) - work done by friction
This is how it makes sense to me.
my mcat book...