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7/23/2019 EGR Homework 1
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Introduction to Engineering
EGR 1301 Section 6
Homework Assignment 1
Robin Vo
Baylor ID 892008866
16 September 2015
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Robin VoEGR 1301-
06
Problem 4.1
The Cartesian components of a vectorB are shown in Fig. 4.5. If Bx, = 8.6 m
andΔ= 35°, findα, By andB.
Assume∠α+Δis a right angle.
α = 90°− ΔwhereΔ is 35°
α = 90°− 35°
α = 55°
TanΔ =By
B x
By = Bx(tanΔ)
By= 8.6 x Tan35°
By= 6.02
CosΔ=By
B
B = Bycos Δ
= 8.6mcos35
B = 10.49 m
Conclusion: B = 10.49m, By = 6.02 m, andα = 55°
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Problem 4.8
A pilot of an aircraft knows that the vehicle in landing configuration will glide
2.0 x 101 km from a height of 2.00 x 10
3 m. A TV transmitting tower is located
in a direct line with the local runway. lf the pilot glides over the tower with
3.0 x 101 m to spare and touches down on the runway at a point 6.5 km from
the base of the tower, how high is the tower?
Convert all measurements to one base unit (meters)
*Note: Not Drawn toScale
20 x 103m
30 m
6500 m
3
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06
Plane glide distance = 2.0 x 101km (
1000m
1km ¿ = 20 x 10
3m
Plane height = 2.00 x 103m
Extra air space = 3.0 x 10
1
mDistance from base of tower = 6.5km (
1000m
1km) = 6500 m or 6.5 x 10
3m
Ratio of glide distance : Height of plane = Distance from where plane lands :
Height of tower
2.0 x 103m : 20 x 10
3m = 6.5 x 10
3m : h+30m, where h is the height of the
tower
Ratio of glide distance : Height of plane = 1:10
Use the known ratio (1
10¿ to find height of tower.
6500m (1
10¿ = 650m − 30m (spare space) = 620m
Conclusion: The height of the tower is 0.62 km.
Problem 4.16
A block of metal has a 90° notch cut from its lower surface. The notched part
rests on a circular cylinder of diameter 2.0 cm, as shown in Fig. 4.15. If the
lower surface of the part is 1.3 cm above the base plane, how deep is the notch
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∠ BAC = 90°
O is the center,∴OB and OC = 1cm since diameter is 2cm , radius has to be
1cm
∠ ABO = ∠ ACO=¿ 90°
Δ ABO andΔ ACO are isosceles, each with 2 legs that are 1cm each,∴ the
remaining leg must be 1 x √ 2 .
AO = √ 2 cm
AX= √ 2 + 1(radius)
Depth of notch= √ 2 + 1 – 1.3 cm = 1.1cm
Conclusion:The depth of the notch is approximately 1.1 cm.
!
"#$
%
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Problem 4.18
An aircraft moves through the air with a relative velocity of 3.00 x 102
km/ hat a heading of N30°E. In a 35 km/h wind from the west,
(a) Calculate the true ground speed and heading of the aircraft.
(b) What heading should the pilot fly so that the true heading is N30°E?N
&
!
'
60°
300(m)r
30°
%++l, law o- co.ine. to /n tre ron .+ee !4
!2= 300x1024235272300x1024354co.60°4
!2= 81225710500!2=9025!= 291 (m)r
;re eain: sin 60°284.1 km/hr
= sinθ300km /hr
Sin60 ° 300(m)r4 = 291 Sin θ¿
25891 = 291 Sin θ¿
0815= Sin θ Sin 7108154= 66 °
Conclusion: ;e ron .+ee o- te +lane i. 291 (m)r an te tre
eain i. N66°E
35 (m)r
S
<
6
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Robin VoEGR 1301-
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Problem 4.30
The light striking a pane of glass is refracted as shown in Fig. 4.20. The law
of refraction states that nasin&a= nb sin&b, where na and nb are the refractive
indexes of the materials through which the light is passing and the angles are
from a Line that is normal to the surface. The refractive index of air is 1.00.
What is the refractive index of the glass?
Apply Snell’s law to find the refractive index of the glass.
Snell’s law: nasin&a= nb sin&b
1.00sin(38 ° ) = nb sin(28 °¿
nb= sin38 °
sin28 °
nb= 1.57
Conclusion:The refraction index of the glass is 1.57.
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Problem 5.8
Electrical resistance for a given material can be a function of both area/unit
thickness and material temperature. Holding temperature constant, a range
of areas are tested to determine resistance. The measured resistance recorded
in the table is expressed in milliohms per meter of conductor length.
(e) Plot and find equations for parts (a), (b), and (c) using computer-assisted
methods.
(f) What would be the best curve fit for this application?
Conclusion:E) For part A,B, and C the equation for the line would be
y = 122.91e-145x
F) The best curve fit for this application would be an exponential curve as the
curve with the plots shows that there is a trend of the line approaching 0.
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Problem 5.18
The rate of absorption of radiation by metal plates varies with the plate
thickness and the nature of the source of radiation. A test was conducted at
Ames Labs on October 11, 2005, using a Geiger counter and a constant source
of radiation; the results are shown in the following table.
C) Repeat parts (a) and (b) using a computer-assisted method.
D) What would you expect the counts per second to be for a 2-in.-thick plate of
the metal used in the test?
C)
Rate of Absorption for Dierent Plate Tic!ness
Equation: y = -144.88x + 4677
D) Convert inches to millimeters
2.00 in (2.54 cm
1 inch)(
10mm
1cm) = 50.8mm
y = -144.88x+4677
y= -144.88(50.8)+4677
y = -2682.904
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Conclusion:The expected count would be 0, since you cannot count the
negative count.
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