Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence
![T_{BE}=W=20*9.81=196.2 N](https://tex.z-dn.net/?f=T_%7BBE%7D%3DW%3D20%2A9.81%3D196.2%20N)
Therefore, tension in the cable, ![T_{BE}=196.2 N](https://tex.z-dn.net/?f=T_%7BBE%7D%3D196.2%20N)
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then
![196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0](https://tex.z-dn.net/?f=196.2%5Ctimes%200.125-%20196.2%5Ctimes%200.2%2B%20D_x%5Ctimes%200.2%3D0)
![24.525-39.24+0.2D_x=0](https://tex.z-dn.net/?f=24.525-39.24%2B0.2D_x%3D0)
![D_x=73.575 N](https://tex.z-dn.net/?f=D_x%3D73.575%20N)
Similarly,
![A_x-D_y=0](https://tex.z-dn.net/?f=A_x-D_y%3D0)
![A_x=73.575 N](https://tex.z-dn.net/?f=A_x%3D73.575%20N)
Therefore, both reactions at A and D are 73.575 N
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
Answer:
W = 2695.14 lb
Explanation:
given,
side of cube = 3 ft
density of the cube = 3.1 slugs/ft³
we know,
![density = \dfrac{mass}{volume}](https://tex.z-dn.net/?f=density%20%3D%20%5Cdfrac%7Bmass%7D%7Bvolume%7D)
mass = density x volume
volume = 3³ = 27 ft³
mass = 3.1 x 27
m = 83.7 slugs.
weight calculation
converting mass from slug to pound
weight of 1 slug is equal to 32.2 lb
now,
weight of the cube is equal to
W = 83.7 slugs x 32.2 lb/slug
W = 2695.14 lb
hence, weight is equal to W = 2695.14 lb