Answer:
The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region
Explanation:
The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
![\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }](https://tex.z-dn.net/?f=%5Cdfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cleft%20%28%5Cdfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%20%20%5Cright%20%29%5E%7B%5Cfrac%7BK-1%7D%7Bk%7D%20%7D)
![{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K](https://tex.z-dn.net/?f=%7BT_%7B2%7D%7D%7B%7D%20%3D%20T_%7B1%7D%20%5Ctimes%20%5Cleft%20%28%5Cdfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%20%20%5Cright%20%29%5E%7B%5Cfrac%7BK-1%7D%7Bk%7D%20%7D%20%3D%20280.15%20%5Ctimes%20%5Cleft%20%289%20%20%5Cright%20%29%5E%7B%5Cfrac%7B1.333-1%7D%7B1.333%7D%20%7D%20%3D%20485.03%5C%20K)
The heat supplied =
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied =
·
= 20 kg/s
The heat supplied = 20*
= 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
![\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}](https://tex.z-dn.net/?f=%5Cdfrac%7BT_5%7D%7BT_4%7D%20%20%3D%20%5Cdfrac%7B2%7D%7B1.333%20%2B%201%7D)
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust =
×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
GPS device details are given below.
Explanation:
Even a simple GPS unit has a wide range of settings and features. Because every unit’s operation varies, this article won’t provide step-by-step details. Read the owner's manual to familiarize yourself with it..
If you’d like additional help, you can also sign up for a GPS navigation class at an REI store.
Though steps vary, all GPS receivers do the following basic functions:
Display position: A GPS tells you where you are by displaying your coordinates; it also shows your position on its base map or topo map.
Record tracks: When tracking is turned on, a GPS automatically lays down digital bread crumbs, called “track points,” at regular intervals. You use those later to retrace your steps or to evaluate the path you traveled.
Navigate point-to-point: A GPS directs you by giving you the direction and distance to a location, or “waypoint.” You can pre-mark waypoints by entering their coordinates at home. In the field you can have the unit mark a waypoint at a place you'd like to return to, such as the trailhead or your campsite. A GPS unit provides the bearing and distance “as the crow flies” to a waypoint. Because trails don’t follow a straight line, the bearing changes as you hike. The distance to travel also changes (decreasing, unless you’re heading the wrong direction) as you approach your goal.
Display trip data: This odometer-like function tells you cumulative stats like how far you’ve come and how high you’ve climbed.
GPS and your computer: GPS units come with a powerful software program that lets you manage maps, plan routes, analyze trips and more. Invest the time to learn it and to practice using all of its capabilities.
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area = ![\frac{\pi d^{2} }{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20d%5E%7B2%7D%20%7D%7B4%7D)
d = 1.5
![\frac{\pi *1.5^{2} }{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%2A1.5%5E%7B2%7D%20%7D%7B4%7D)
= 1.767 inches²
<h3>Solve for inertia</h3>
![\frac{\pi *0.75^4}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%2A0.75%5E4%7D%7B4%7D)
= 0.2485inches⁴
Solve for the tensile force from here
![\frac{F}{A} +\frac{Mc}{I}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%2B%5Cfrac%7BMc%7D%7BI%7D)
30x10³ = ![\frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\](https://tex.z-dn.net/?f=%5Cfrac%7BP%7D%7B1.767%7D%20%2B%5Cfrac%7B4.75p%2A0.75%7D%7B0.2485%7D%20%5C%5C%5C%5C)
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
Read more on tensile force here: brainly.com/question/25748369