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:
Change in length = 0.0913 in
Explanation:
Given data:
Length = 6 ft
Diameter = 0.2 in
Load w = 200 lb/ft
Solution:
We start by applying the equilibrium moment about point C, expressed as
∑M(c) = 0.............1
This can be used to find the force in AB.
10× 200 × ( 5) - (T cos(30)) × 10 = 0
Solving gives us
Tension in wire T(AB) = 1154.7 lb
We also know the modulus of elasticity for A992 is
E = 29000 ksi
And the area will be
Area = 
The change in length is expressed as
Change in length = 
.........2
Substituting values results in
Change in length = 
Change in length = 0.0913 in
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:
Cite two significant refinements that emerged from the Wave-mechanical
atomic model.
Answer:
Provided information
an electronic system dissipating = 90 W
diameter = 15 cm
The duct has components cooled by forced air entering at 32°C with a flow rate of 0.65 m3 /min
85% of the generated heat is transferred to the air in the duct, and 15% is lost through the duct's outer surfaces.
See the attached images for the solution.
Explanation:
Consult the attached pictures for a thorough breakdown of the explanation.
