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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Complete the following problems. Show your work/calculations, save as.pdf and upload to the
assignment in Blackboard.
missing information to present a completed program. (Hint: You may have to look up geometry
for the center drill and standard 0.5000 in twist drill to know the required depth to drill).
1. What are the x and y dimensions for the center position of holes 1,2, and 3 in the part shown in
Figure 26.2 (below)?
6.0000
Zero
reference
point
7118
1.0005
1.0000
1.252
Bore
6.0000
.7118
Cbore
0.2180 deep
(3 holes)
2.6563 1.9445
Figure 26.2
026022 (8lot and Drill Part)
(Setup Instructions---
(UNITS: Inches
(WORKPIECE NAT'L SAE 1020 STEEL
(Workpiece: 3.25 x 2.00 x0.75 in. Plate
(PRZ Location 054:
'
XY 0.0 - Upper Left of Fixture
TOP OF PART 2-0
(Tool List
( T02 0.500 IN 4 FLUTE FLAT END MILL
#4 CENTER DRILL
Dashed line indicates-
corner of original stock
( T04
T02
3.000 diam. slot
0.3000 deep.
0.3000 wide
Intended toolpath-tangent-
arc entry and exit sized to
programmer's judgment…
A program to make the part depicted in Figure 26.A has been created, presented in figure 26.B, but some information still needs to be filled in. Compute the tool locations, depths, and other missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill).
We consider a laminar flow induced by an impulsively started infinite flat
plate. The y-axis is normal to the plate. The x- and z-axes form a plane parallel
to the plate. The plate is defined by y = 0. For time t <0, the plate and the flow
are at rest. For t≥0, the velocity of the plate is parallel to the 2-coordinate;
its value is constant and equal to uw. At infinity, the flow is at rest. The flow
induced by the motion of the plate is independent of z.
(a) From the continuity equation, show that v=0 everywhere in the flow and
the resulting momentum equation is
მu
Ət
Note that this equation has the form of a diffusion equation (the same form as
the heat equation).
(b) We introduce the new variables T, Y and U such that
T=kt, Y=k/2y, U = u
where k is an arbitrary constant. In the new system of variables, the solution
is U(Y,T). The solution U(Y,T) is expressed by a function of Y and T and the
solution u(y, t) is expressed by a function of y and t. Show that the functions are
identical.…
Java How to Program, Early Objects (11th Edition) (Deitel: How to Program)
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