Response:
a) 4 kg/s
b) 99.61 °C
Rationale:
Refer to the pictures provided.
Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,
spherical container 1.1 m in diameter. The opaque, diffuse, gray container surfaces have an emissivity of 0.05 and are separated by an evacuated space.
a. If the outer surface is at 280 K and the inner surface is at 90 K, what is the mass rate of oxygen lost due to evaporation?
a. If the outer surface is at 280 K and the inner surface is at 90 K, what is the mass rate of oxygen lost due to evaporation?
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Answer:
Refer to the attached document
1512 ft
Explanation:
Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:
Acceleration - Time
0 < t < 6: Velocity change = area beneath the a–t graph
V_6 - 0 = (6 s)(4 ft/s²) = 24 ft/s
6 < t < t2: The velocity rises from 24 to 48 ft/s,
Velocity change = area beneath the a–t graph
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: The velocity remains constant, meaning acceleration is zero.
34 < t < 40: Velocity change = area beneath the a–t graph
0 - 42 = 6*a4
a4 = - 8 ft / s²
A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.
Velocity - Time
Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.
Position change = area beneath the v−t graph
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Summing these position changes yields the distance from A to B:
d = x40 - 0 = 1512 ft
Answer:
a. 25! = (Approximately)
b. 24!
Explanation:
a. In a Playfair cipher, there are 25 keys available because it is structured in a 5 * 4 grid. By using permutations to enumerate all potential configurations, we derive: 25! = 1.551121004×10²⁵ =
Although there are 26 letters available, in the Playfair cipher, the letters 'i' and 'j' are treated as a single letter.
b. Considering any configuration of 5x5, each of the four row shifts yields equivalent configurations, amounting to five total equivalencies. Similarly, for each of these five setups, any of the four column shifts also results in equivalent arrangements. Therefore, each configuration corresponds to 25 equivalent arrangements. Consequently, the total count of distinct keys can be expressed as:
25!/25 = 24! = 6.204484017×10²³