37. Which forms of matter may carry mechanical waves? gases only gases and liqui
ID: 1397614 • Letter: 3
Question
37.
Which forms of matter may carry mechanical waves?
gases only
gases and liquids only
solids, liquids, and gases
liquids and solids only
38.
A cannon of mass 1200 kg is positioned at the edge of a cliff, where it fires a 100.0 kg cannonball horizontally. The cannonball's initial speed is 35 m/s . What is the recoil speed of the cannon? Assume that friction forces are negligible.
2.9 m/s
35 m/s
1.6 m/s
3.5 m/s
39.
As a roller coaster car travels up and down a trackway, its total mechanical energy is _____.
transformed to kinetic energy only
gradually dissipated by friction
transformed between potential and kinetic energy, but no other forms
transformed to potential energy only
40.
A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep the box moving at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?
0.23
0.17
0.20
0.18
37.
Which forms of matter may carry mechanical waves?
gases only
gases and liquids only
solids, liquids, and gases
liquids and solids only
38.
A cannon of mass 1200 kg is positioned at the edge of a cliff, where it fires a 100.0 kg cannonball horizontally. The cannonball's initial speed is 35 m/s . What is the recoil speed of the cannon? Assume that friction forces are negligible.
2.9 m/s
35 m/s
1.6 m/s
3.5 m/s
Explanation / Answer
mechanical waves requires medium for their propogation
so they can travel in solids, liquids, and gases
++++++++
brfore firing the cannon and cannon ball is at rest v1 = v2 = = 0
initial momentum Pi = m1*v1 + m2*v2 = 0
after firing speed of cannon ball = v2' = 35 m/s
speed of cannon = v1'
final momentum Pf = m1*v1' + m2*v2'
from momentum conservation initial momentum = final momentum
m1*v1' + m2*v2' = 0
1200*v1' + 100*35 = 0
v1' = 3500/1200 = 2.91 m/s <-----------------answer
+++++++++++
transformed between potential and kinetic energy, but no other forms
+++++++
here normal force = N = m*g = 50*9.8 = 490 N
frictional force f = u*N = u*490
u = coeffiient of friction
applied force Fapp = 99 N
net force Fnet = Faap - f
from newtons law
Fnet = m*a
here speed is constant aceleration = a= 0
Fapp - f = 0
Fapp = f
u = 99/490
u = 0.20
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