The value of the gravitational acceleration g decreases with elevation from 9.807 m/s 2 at sea level to 9.767 m/s 2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level.
The value of the gravitational acceleration g decreases with elevation from 9.807 m/s 2 at sea level to 9.767 m/s 2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level.
The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at sea level to 9.767 m/s2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level.
Expert Solution & Answer
To determine
The reason why a bicyclist on a downhill road picks up speed even when he is not pedaling and also find whether this violates the principle of conservation of energy?
Explanation of Solution
In this case, there are two forms of energy involved; potential energy and kinetic energy. When the bicyclist is at the top of a hill, he has potential energy due to his mass. When bicyclist moves, the energy changes its form to kinetic energy due to his motion. Thus, the bicyclist picks speed when he moves downhill even when he is not pedaling.
There is a transformation in the forms of energy which states the principle of energy. Hence, there is no violation of the principle of conservation of energy.
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For the walking-beam mechanism shown in Figure 3, find and plot the x and y coordinates of the
position of the coupler point P for one complete revolution of the crank O2A. Use the coordinate
system shown in Figure 3. Hint: Calculate them first with respect to the ground link 0204 and
then transform them into the global XY coordinate system.
y
-1.75
Ꮎ
Ꮎ
4
= 2.33
0242.22
L4
x
AP = 3.06
L2 = 1.0
W2
31°
B
03 L3 = 2.06
P
1
8
5
.06
6
7
P'
The link lengths, gear ratio (2), phase angle (Ø), and the value of 02 for some geared five bar
linkages are defined in Table 2. The linkage configuration and terminology are shown in Figure
2. For the rows assigned, find all possible solutions for angles 03 and 04 by the vector loop
method. Show your work in details: vector loop, vector equations, solution procedure.
Table 2
Row
Link 1 Link 2
Link 3
Link 4
Link 5
λ
Φ
Ө
a
6
1
7
9
4
2
30°
60°
P
y 4
YA
B
b
R4
R3
YA
A
Gear ratio:
a
02
d
05
r5
R5
R2
Phase angle: = 0₂-202
R1
05
02
r2
Figure 2.
04
X
Problem 4
A .025 lb bullet C is fired at end B of the 15-lb slender bar AB. The
bar is initially at rest, and the initial velocity of the bullet is 1500 ft/s
as shown. Assuming that the bullet becomes embedded in the bar,
find (a) the angular velocity @2 of the bar immediately after impact,
and (b) the percentage loss of kinetic energy as a result of the impact.
(c) After the impact, does the bar swing up 90° and reach the
horizontal? If it does, what is its angular velocity at this point?
Answers: (a). @2=1.6 rad/s; (b). 99.6% loss
=
(c). Ah2 0.212 ft. The bar does not reach horizontal.
y
X
4 ft
15 lb
V₁
1500 ft/s
0.025 lb
C
30°7
B
A
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