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Unformatted text preview: gravity
mg = ρV g = A(0.5 m)g downward, and the buoyancy force
Fb = ρf Vd g = ρf (A(0.3 m))g upward. Thus at equilibrium
ρA(0.5)g = ρf A(0.3)g =⇒ ρ = (1000 kg/m3 )
3 (b) 0.3
= 600 kg/m3
0.5 B Suppose I use the same wood to make a block which is
twice as wide. Which of the following pictures shows the way this
new block will ﬂoat? (A)
same as
above (B) 0.5m (C) 1m
The level at which a block ﬂoats depends on density, not
mass or volume. Page 10 21. Water ﬂows through a pipe which widens from a crosssectional area of AL = 0.1 m3 to a crosssectional area
of AR = 0.4 m3 . The water enters the pipe moving at
vL = 4 m/s. The pressure of the water as it leaves the
pipe is PR = 105 Pa.
3 (a) A
A) 1 m/s AR=0.4m3
AL=0.1m3
vL =
4 m/s PR =
105 Pa The speed of the water vR as it leaves the pipe is
B) 2 m/s C) 4 m/s D) 8 m/s E) 16 m/s The ﬂux Φ = Av must be the same at both ends. 3 (b) Find the pressure PL of the water as it enters the pipe, ignoring the eﬀects of viscosity. 12
12
PL + ρvL = PR + ρvR
2
2
12
2
PL = PR + ρ(vR − vL )
2
1
= 105 Pa + (1000 kg/m3 )((1 m/s)2 − (4 m/s)2 )
2
= 105 Pa − 7500 Pa
= 92500 Pa
3 22. Answer one of these two questions, or both for extra credit.
(a) B Planet X has twice the radius of Earth and twice the mass of Earth. What is the acceleration due to gravity g on Planet X?
A) 2.5 m/s2 B) 4.9 m/s2 C) 9.8 m/s2 D) 19.6 m/s2 E) 39.2 m/s2 M
(so that
R2
F = mg ). If we double M and double R, we end up with half the g we had before. The force on an object on the surface of a planet is F = G M m , so g = G
R2 (b) A satellite orbits the Earth at an altitude of r above the Earth’s center. Write an expression for the
satellite’s velocity v , in terms of the Earth’s mass M . The Earth’s gravity is the centripetal force which keeps the satellite orbiting:
Mm
mv 2
=
r2
r
GM
=⇒ v 2 =
=⇒ v =
r
G GM
r which is about 70% of the escape velocity at that altitude (by the way). Page 11 3 23. Extra credit: A 5 kg wheel with a radius of 2 m and moment
1
of inertia I = 2 M R2 is mounted on an axle and can spin without
friction. It starts at rest, and then a 30 N force is applied to the
top of the wheel. What is the angular velocity ω of the disk after
the disk has made one complete revolution? 30N 5kg
There are several ways to do this: by calculating the angular acceleration α and using kinematics, or by using angular
momentum. I’m going to use energy here. The torque exerted on the wheel is τ = rF = (2 m)(30 N) = 60 N · m,
and it is applied over one complete revolution, or an angular
displacement ∆θ = 2π . Thus the work done by the force is
W = F ∆θ = (60 N · m)(2π ) = 120π
This work is energy which goes into kinetic energy, and so
after one revolution
120π J =
=
=
=⇒ ω 2 =
=⇒ ω = 12
Iω
2
11
( M R2 )ω 2
22
1
(5 kg)(2 m)2 ω 2
4
120π
= 24π
5
√
24π = 8.7 rad/s Page 12 2m...
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This note was uploaded on 01/15/2014 for the course PHYSICS 2130 taught by Professor Scotthill during the Spring '12 term at Toledo.
 Spring '12
 ScottHill
 Physics, Acceleration

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