panesar (rsp739) HW02 gilbert (54345)
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001 (part 1 of 2) 10.0
panesar (rsp739) HW03 gilbert (54345)
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001
1
For then the poi
baek (mjb3994) HW15 goddard (54330)
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001
002
1
10.0 points
Eva
baek (mjb3994) HW03 goddard (54330)
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001
keywords: determinan
baek (mjb3994) HW13 goddard (54330)
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10.0 points
1
Since
Version 030 TEST04 tsoi (91875)
What is the wavelength at any position in
front of the ambulance for the sound from the
ambulances siren? The velocity of sound in
air is 343 m/s.
1. 0.583661
2. 0.4591
Version 013 TEST03 tsoi (91875)
After the collisions the rod and clay system
has an angular velocity about the pivot.
2
Correct answer: 5.20354 m/s .
Explanation:
Let : M = 5.3 kg ,
r = 0.634 m ,
m =
6
2.7 m
Version 013 TEST03 tsoi (91875)
45 cm
3. 8.28251
4. 6.71416
M
5. 8.63018
6. 9.49526
7. 9.18041
8. 7.27461
26
9. 9.28655
10. 5.77235
Find the speed of the sphere when it reaches
the bottom of t
Version 013 TEST03 tsoi (91875)
m2 = 17 kg ,
h = 1.6 m .
and
The wheel starts from rest, so
From conservation of energy
K1 + K2 + Kdisk = U
m1
m2 v 2 M v 2
h
+
+
= (m1 m2 ) g
2
2
4
2
2
(2 m1 + 2 m2 +
Version 013 TEST03 tsoi (91875)
The lever rod is in equilibrium at angle of 74
from the vertical wall. The cable makes angle
of 50 with the rod.
What is the tension of the supporting cable?
1. 47.8934
Version 030 TEST04 tsoi (91875)
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001 10.0 points
Earthquakes
Version 030 TEST04 tsoi (91875)
antinodes and 8 nodes: one node at each end,
and 6 in the middle.
007 10.0 points
A block of unknown mass is attached to a
spring of spring constant 9.2 N/m and undergo
Version 013 TEST03 tsoi (91875)
The pivot provides an external force to the
rod+clay system.
III) Angular momentum is conserved, since
there is no external torque applied to the
rod+clay system.
2.4 m
Version 013 TEST03 tsoi (91875)
Nw
mg
f
P ivot
Nf
Fx = f Nw = 0
f = Nw
Fy = Nf m g = 0
= m g d cos Nw L sin = 0
Nw L sin = m g d cos
f L sin = m g d cos
mgd
f=
L tan
The ladder may slip when f = f
Version 030 TEST04 tsoi (91875)
The tension in the string is
2. y = A sin
F = mg
and the mass per unit length of the string is
v=
=
=
F
4. y = A sin
F
mg
= 2
2
v
v
(3.8 kg) 9.8 m/s2
2
(20 m/s)
5. y =
Version 030 TEST04 tsoi (91875)
On a standing wave, the distance of the nearest two nodes is
d=
= ,
2
k
where is the wavelength. In this problem,
x = 0 is a node, which can be seen by adding
y1 and y2
Version 036 TEST01 tsoi (91875)
the path and points P and R are the same
height above the ground.
Q
P
R
3
Explanation:
Since air friction is negligible, the only acceleration on the ball after being t
Version 013 TEST03 tsoi (91875)
6 kg
= (6 kg + 6 kg + 6 kg + 2 kg)
(4 m)2 + (6 m)2
4
2 kg
y
6m
= 260 kg m2 .
O
x
6 kg
6 kg
4m
If the system rotates in the xy plane about
the center of the rectangle (z
1.3 Accumulation and Amount Functions
Suppose we open a savings account with an amount $K .
We will say that the day we open our savings account
corresponds to time t = 0 . We refer to the original
in
Version 036 TEST01 tsoi (91875)
7
y0 = 0 m .
y
t
2.
y
y = y0 + v0 t +
t
3.
This is a parabolic shaped curve starting at
y = 0 with a continuously decreasing slope as
time increases.
y
t
4.
y
t
5.
corr
Version 036 TEST01 tsoi (91875)
from a battleship as v0 . When the initial
projectile angle is 45 with respect to the
horizontal, it gives a maximum range R.
y
45
x
R/2
R
The time of ight tR of the ca
Version 036 TEST01 tsoi (91875)
The acceleration of gravity is 10 m/s2 . Air
resistance is negligible.
A student rolls a 11 kg ball o the horizontal
roof of the building in the direction of the
target
Version 022 TEST02 tsoi (91875)
The total energy is
1. 1421.62
2. 1720.39
3. 1585.64
4. 1525.03
5. 1692.9
6. 1472.74
7. 1546.66
8. 1398.56
9. 1662.11
10. 1628.86
1
GmM
m v2
.
2
d
1 GM
GmM
= m
2
d
d
G
Version 022 TEST02 tsoi (91875)
after
From conservation of momentum,
m1 v1i + m2 v2i = m1 v1f + m2 v2f
Y
2.
6 m/s
after
correct
3.
37
Y
8 m/s
m v = m v1 + m v2
v = v1 + v2 .
This relation leads to the
Version 036 TEST01 tsoi (91875)
Given : W1
W2
1
3
T2
013 10.0 points
A car makes a 256 km trip at an average
speed of 39 km/h. A second car starting 1 h
later arrives at their mutual destination at th
Version 022 TEST02 tsoi (91875)
This print-out should have 14 questions.
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before answering.
1
2. 6
Mg
=
0
21
001 10.0 po
Version 022 TEST02 tsoi (91875)
The two blocks are connected by a light
string that passes over a frictionless pulley
with a negligible mass. The 4 kg block lies
on a rough horizontal surface with a c
September 28
Relevant reading: Section 2.1, 2.2 and 2.4
Consider a
nonhomogeneous
linear second order ODE
d2 y
dy
+ q (t) y = g (t)
+ p (t)
2
dt
dt
We know how to solve the homogeneous case
g0
(1)
in
September 26
Relevant reading: Section 2.1 and 2.2
We are going to show how to nd a second solution y2 of the homogeneous
second order ODE
dy
d2 y
+ q (t) y = 0
+ p (t)
2
dt
dt
given a rst solution y1
M 427J (Unique 54167) - Fall 2017 - Dierential Equations with Linear
Algebra
Lecture: TTh 9:30 am - 11:00 am, GSB 2.126.
Problem Session: MW 8:00 am - 9:00 am, CLA 0.126.
Instructor: Ryan Denlinger
E-