A spherical hot-air balloon is initially filled with air at 120 kPa and 20°C with an initial diameter of 5 m. Air enters this ba
1 answer:
Answer:
The duration is 17.43 minutes.
Explanation:
Based on the provided information, the initial diameter is 5 m
the velocity is 3 m/s
and the final diameter is 17 m.
To find the solution, we will use the volume change equation expressed as
ΔV = .............1
where ΔV represents the change in volume, rf is the final radius, and ri is the initial radius.
Calculating ΔV yields
ΔV =
ΔV = 2507 m³.
Thus,
Q = velocity × Area
Q = 3 × π ×(0.5)² = 2.356 m³/s.
Next, the change in time can be expressed as
Δt =
Δt =
Δt = 1046 seconds.
Therefore, the total change in time amounts to 17.43 minutes.
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Response:
a) 144.000 seconds
b) and c) Battery voltage and power graphs are in the attached image.
where D:{0<t h="" />
d) 1620 J
Description:
a) The initial response is derived via a rule of three
b) Using the line equation from the starting point (0 seconds, 1.5 V)
where m denotes the slope.
where V represents voltage in volts and t signifies time in seconds
along with P and m.
d) By evaluating that count.
Response:
a) The diameter available is 0.0384 nm
b)This space is less than the size of a carbon atom, which has a radius of 0.077 nm, indicating that the carbon atom won't occupy these sites.
Clarification:
For BCC iron
According to the information in Appendix B, the lattice parameter (a) is determined to be 0.2866 nm
BCC iron encompasses 4 atomic radii, thus the body diagonal length =
which represents the atomic radius of BCC iron
4r =
Substituting the value of (a) from Appendix B, set as 0.2866 nm
4r =
leading to r = 0.4964 nm / 4 = 0.1241 nm
Refer back to Appendix C, where the atomic radius of BCC iron is stated as 0.1241 nm, assuming the atomic sizes for iron remain consistent.
Thus, the radius ratio = 0.62
According to Figure 3.2, the space necessary for an interstitial at the BCC site exists between atoms located at the FCC site, containing two atoms, each equal to a radius of 0.2482 nm
The diameter of the minimum space available
= 0.2482 nm
With a = 0.2666 nm
therefore
d = 0.2866 nm - 0.2482 nm = 0.0384 nm
When comparing with the diameter of a carbon atom
This space is smaller than that of a carbon atom which has a radius of 0.077 nm, confirming that the carbon atom will not be able to occupy these positions.