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Question 1
The length of the airplane is 63 ft., which is equivalent to 19.20 m (Wikipedia contributors, 2024).
Question 2
The diameter of the balloon can be estimated by the shadow cast by the U-2 aircraft. The method works based on:
- As sunlight reaches the earth, the rays are nearly parallel due to the distance between earth and the sun.
- Therefore, as the shadow falls on the balloon, the line which connects the shadow to the aircraft is nearly perpendicular to the airplane. Its long dimension intersects the balloon.
- So, the line is used as a valuable stick to estimate the diameter of the balloon because it represents the actual distance from the aircraft to the shadow.
- Measuring the length of the canopy helps to estimate the diameter of the balloon.
The explanation can be actualized to determine the diameter of the balloon, which is equal to the airplane length of 19.20 m
Question 3
To determine the volume of the balloon, the formula applied to calculate the volume of a sphere = is used. The assumption is based on the information provided about the balloon being nearly spherical. The radius is half of the diameter = 19.20/2 = 9.6 m.
V = = = 3705.97.
Question 4
Inspecting the units on the right-hand side of equation H:
H = KT/mg (Wikipedia contributors, 2023)
Where K is Boltzmann constant =
T is the temperature in Kelvin, K
m is mass in kg
g is the acceleration due to gravity in
In unit form,
Simplifying the equation through cancellation, m remains as the units for H, which is meters.
Question 5
Nitrogen molecule has two nitrogen atoms. Therefore, the mass = 14×2=28.
The atomic mass unit of nitrogen is converted to kg.
One mass atomic unit =. For 28 =.
H = KT/mg
H = = 83268.52m = 83.27km
Question 6
Calculating the atmospheric pressure at Balloon altitude:
P =
The scale height obtained in question 5 H of 83268.52m and an altitude, z of 18,000m are used. The surface pressure is 1013 hpa.
P = =816.27 hpa.
Question 7
Finding the number of molecules at the obtained atmospheric pressure:
Using the ideal gas law: P = nKT.
Therefore, n = P/KT.
n = 81627pa/ = molecules /volume.
Question 8
The mass displaced is calculated using the density of air at that altitude and the volume of the balloon to find molecules displaced. The mass is then gotten from there.
m = n. V where n is molecules /volume, and V is 3705.97.
m = ×3705.97 = molecules
Mass = = 3703.53 kg.
Question 9
To find the lift in newtons, Newton’s second law, F = mg is used.
F = 3703.53 × 9.81 =36331.59 N.
Question 10
The capacity of the balloon to lift a given load diminishes with altitude because as the balloon ascends, the atmospheric pressure decreases. Thus, the air density decreases according to Archimedes’ Principle of diminishing buoyancy force (Poso, 2018). It explains the reason, at higher altitudes, the balloon has a lower lifting force.
Party balloons do not rise to the fringes of outer space because as they rise, they eventually reach an altitude where atmospheric pressure is too low for them to get sufficient lift.
References
Poso, F. J. (2018). Buoyancy. The Archimedes principle. GRIN Verlag.
Wikipedia contributors. (2023, December 12). Scale height. Wikipedia. https://en.wikipedia.org/wiki/Scale_height
Wikipedia contributors. (2024, March 12). Lockheed U-2. Wikipedia. https://en.wikipedia.org/wiki/Lockheed_U-2
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Question 9
- The average elastic modulus of nylon when the forces are low is 0.4 × 106 𝑝𝑠𝑖, which is high.
Since the value represents the stiffness of nylon, the high value means that nylon is quite stiff, exhibiting strong elastic characteristics even under low forces.
- As the forces rise above the 5N level with the nylon filament under load, the following happens:
Initially, under considerable forces, the spring behaves according to Hooke’s law: 𝐹 = −𝐾𝑥 (Huang, 2002). F is spring force, K is spring constant, and x is spring stretch. Thus, there is direct proportionality between strain and the force per unit area.
- Deformation occurs above 5 N. The reason is that it has exceeded the elastic limit. In the case of nylon, the filament deforms permanently once it exceeds the elastic limit.
- The material exhibits a non-linear response to Hooke’s law since the elastic limit is exceeded.
- A practical situation where the non-linear response is apparent is when safety belts are used in vehicles.
- In lower forces, under normal use, the seat belts behave linearly by absorbing energy and protecting passengers. Hence, they obey Hooke’s law.
- However, in a severe crash, the seat belts show non-linear behavior. Due to excess force on the belts, they can experience plastic deformation (Huang, 2002). This minimizes their effectiveness of restraining and holding back the passenger.
- Thus, the seat belts are designed under careful considerations that balance elasticity and strength to ensure passengers are protected effectively.
References
Huang, M. (2002). Vehicle crash mechanics. CRC Press.
