Answer:
Aluminum cross sectional area = 1.99 * 10^-5 mm^2
Steel cross sectional area = 9.95* 10^-6 mm^2
Explanation:
Given data:
Tensile strength of Grunerite = 3.5 * 10^2 kg/mm^2 = 3.5 * 10^-4 kg/<em>u</em>m^2
Tensile strength of Aluminum = 2.5 × 10^4 lb/in2 = 2.5 × 10^4 * 703.07 kg/m^2
Tensile strength of Steel number 5137 = 5.0 × 10^4 lb/in2 = 5.0 × 10^4*703.07 kg/m^2
<u>Calculating the cross sectional area of the wires of aluminum and steel No5137</u>
first we will determine the cross sectional area of the aluminum wire ( A ) by equating tensile strength of aluminum with the tensile strength
of Grunerite
(2.5 × 10^4 * 703.07) * A = 3.5 * 10^-4 kg
Hence ; A = 1.99 * 10^-5 mm^2
Next we calculate the cross sectional area of steel
5.0 × 10^4*703.07 kg/m^2 * A" = 3.5 * 10^-4 kg
Hence ; A" = 9.95* 10^-6 mm^2
Answer
given,
6 lanes divided highway 3 lanes in each direction
rolling terrain
lane width = 10'
shoulder on right = 5'
PHF = 0.9
shoulder on the left direction = 3'
peak hour volume = 3500 veh/hr
large truck = 7 %
tractor trailer = 3 %
speed = 55 mi/h
LOS is determined based on V p
10' lane weight ; f_{Lw}=6.6 mi/h
5' on right ; f_{Lc} = 0.4 mi/hr
3' on left ; no adjustment
3 lanes in each direction f n = 3 mi/h
= 0.877
= 1,555 veh/hr/lane
= (55 + 5) - 6.6 - 0.4 -3 -0
= 50 mi/h
level of service is D using speed flow curves and LOS for basic free moving of vehicle
Answer:
0.5°c
Explanation:
Humidity ratio by mass can be expressed as
the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air
Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.
Humidity ratio expressed by mass:
x = mw / ma (1)
where
x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)
mw = mass of water vapor (kg, lb)
ma = mass of dry air (kg, lb)
It can be as:
x = 0.005 (100) / [(100 - 100)]
x = 0.005 x 100 / (100 - 100)
x = 0.005 x 100 / 0
x = 0.5°c
So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c
Answer:
1. Equatorial Evergreen or Rainforest
2. Tropical forest
3. Mediterranean forest
4. Temperate broad-leaved forest
5. Warm temperate forest
Explanation:
Answer:
See attachment and explanation.
Explanation:
- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.
- The plot of the graph is given in attachment.
- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.
