Answer:
shaft is not transform 100% power
Explanation:
given data
current = 10 amp
voltage = 110 v
torque = 10.2 Nm
rotating speed N = 1000 rpm
solution
we get here first power that is consume by motor is express as
power = current × voltage .........1
power = 10 × 110
power = 1100 W
and power that transmit by shaft is express as
power = T × ω
power = T ×
power = 10.2 ×
power = 1068.14 W
so we can say from above both power
shaft is not transform 100% power
Hai
Your answer will be A.
If you lower the Air Pressure your Object will Float Down ward. The Air Pressure allows it to Fly.
Answer:
The distinguished ambiguities in a reasonable manner are following.
- In what limit will the robot choose when to snap a photo
- Where will the photographs be put away
- What is the memory capacities of the limit office
- Will the machine have the ability to take pictures around evening time
- Can the ground overseer change the route in the midst of flight
- What will happen if the robot can't locate the principal route after avoidance of impediments
- Can more than one objective be transferred
- What will happen when the goal is found
- Is a zone gotten a good deal on where the goal is found
- In what manner will the robot prompt the ground director that the goal is found
- Can the ground director request the machine to return before the preset way is done
The above portrayal utilizing the organized methodology depicted right now following
Capacity:Search and Recovery Drone
Depiction:Drone used in the midst of request and recovery to find the goal on a set way
Info:Target picture and set way
Source:Ground director
Yields High:goals pictures composed to target
<u>Objective </u>
Activity:The robot fly as demonstrated by the preset way and take significant standards pictures until the goal is composed to the photos taken
Requires: Preset way and target picture
Precondition:Target picture must match pictures taken
Post condition :Target is found
Reactions:Target isn't found or composed
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress ![\sigma = \dfrac{pd}{2t}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7Bpd%7D%7B2t%7D)
Making thickness t the subject; we have
![t = \dfrac{pd}{2* \sigma}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7Bpd%7D%7B2%2A%20%5Csigma%7D)
![t = \dfrac{560000*3}{2*150000000}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B560000%2A3%7D%7B2%2A150000000%7D)
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
![\sigma = \dfrac{pd}{4t}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7Bpd%7D%7B4t%7D)
![t= \dfrac{pd}{4*\sigma }](https://tex.z-dn.net/?f=t%3D%20%5Cdfrac%7Bpd%7D%7B4%2A%5Csigma%20%7D)
![t = \dfrac{560000*3}{4*150000000}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B560000%2A3%7D%7B4%2A150000000%7D)
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm