A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i
n. wide by 12 in. deep. The beam carries a 9-kip concentrated load at midspan. Which point has the largest Q value at section a–a?
1 answer:
Point B is where 
shows the highest Q value at section a–a. The missing diagram that was supposed to accompany this question can be found in the attached file. Based on the given information, we must identify the point with the greatest Q value at section a–a. To achieve this, we're working with the attached image. The image reveals that there are 8 blocks stacked vertically on the right-hand side, totaling 12 inches; thus, each block measures 1.5 inches in height. Additionally, the blocks are divided into sections marked as A, B, C, and D. For point A, we use the formula Q representing the moment of Area A. For point B, we set Q as the moment of Area B, and similarly for points C and D. However, point D has no area, resulting in a moment of 0.
You might be interested in
Answer:
Explanation:
The half-life for the specified RC circuit can be expressed as
where [/tex]\tau = RC[/tex]
Given
The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.
Thus, the new half-life is
Now, divide equation 2 by 1
After substituting all values, we can calculate the revised half-life
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
Answer:
a) , b)
Explanation:
a) The uniform dresser can be modeled using specific equilibrium equations:
Following some algebraic manipulations, the formulated equation is derived:
b) Similarly, the man can be represented by a set of equilibrium equations:
After some algebraic changes, the expression for the coefficient of static friction comes out as: