V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
Answer:
m v1 = (m + M) v2
v2 = m v1 / (m + M)
v2 = 7 * 74 / (74 + 65)
3.73 m/s
74 kg is too heavy for the cannonball (over 150 lbs)
Answer:
The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².
Explanation:
By Hooke's law,
F is the force on a spring, k is the spring constant and e is the extension or compression.
From the question,
This is the force on the mass suspended on the spring. Its acceleration, a, is given by
This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of
12.96 - 9.8 m/s² = 3.16 m/s²
Answer:5.7m/s
Explanation:
Mass=1kg
Initial velocity=u=8m/s
height=h=1.6m
Final velocity =v
Acceleration due to gravity=g=9.8m/s^2
v^2=u^2-2xgxh
v^2=8^2-2x9.8x1.6
v^2=8x8-2x9.8x1.6
v^2=64-31.36
v^2=32.64
Take the square root of both sides
√(v^2)=√(32.64)
v=5.7
Speed at the height of 1.6m is 5.7m/s
You get a more low sound.
Conversely, when the wavelength becomes shorter you get a more treble sound.
;-)