Answer: 1766.667 Ω = 1.767kΩ
Explanation:
V=iR
where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).
3mA = 0.003 A
Rearranging the equation, we get
R=V/i
Now we are solving for resistance. Plug in 0.003 A and 5.3 V.
R = 5.3 / 0.003
= 1766.6667 Ω
= 1.7666667 kΩ
The 6s are repeating so round off to whichever value you need for exactness.
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

Explanation:
<h2>Given:</h2>








<h2>Required:</h2>
Distance from Station A to Station B

<h2>Equation:</h2>




<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>
Solve for 





Solve for 




Solve for 





<h3>Distance when a = 0 m/s²</h3>



Solve for 





Solve for 




Solve for 





<h3>Distance when a = -0.8 m/s²</h3>



Solve for 






Solve for 




Solve for 





<h3>Total Distance from Station A to Station B</h3>





<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>
Answer:
Explanation:
Floor Load:
Lo= 50psf
At= 25x25 = 625 square feet

= 13.1psf
%reduction= 13.1/50 = 26%
Fr= 3[(13.1psf)(25ft)(25ft)+(20psf)(25ft)(25ft)]= 62k