Answer:
hello the required diagram is missing attached to the answer is the required diagram
7.9954 kip.ft
Explanation:
AB = 1550-Ib ( weight acting on AB )
BCD = 190 - Ib ( weight of cage )
169-Ib = weight of man inside cage
Attached is the free hand diagram of the question
calculate distance 
= cos 75⁰ = 
= 2.59 ft
calculate distance x
= cos 75⁰ = 
x = 30 * cos 75⁰ = 7.765 ft
The resultant moment produced by all the weights about point A
∑ Ma = 0
Ma = 1550 * + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )
Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )
= 4014.5 + 1950.35 + 2030.535
= 7995.385 ft. Ib ≈ 7.9954 kip.ft
Answer:
0.31 μm
Explanation:
this question wants us to Determine the depletion region width, xn, in the n-side in unit of μm. using the information below.
density in the p-side = 5.68x10^16
density in the n-side = 1.42x10^16
= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)
= √150.74x10⁻¹¹
= 3.882x10⁻⁵
approximately 0.39μm
xn = 0.39 x 0.8
= 0.31μm
0.31 um is the depletion region width. thank you!
Answer:
hello your question is incomplete attached below is the complete question
answer : Drag force = 1.3 Ib
Explanation:
we have to represent the dimensions of the drag force in terms of FLT
i.e : D = f( <em>d,v,p,u </em>) represented in terms of FLT
D = F , V = LT^-1, d = L, p = FL^-4 T^2
u = FL^-2 T, Number of independent terms = 5
attached below is the detailed solution
Answer:
2132hp ed enregia
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Explanation:
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Answer:
well what do you wanna talk about friend?
Explanation:
