Answer:
note:
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Answer:
A) About
newtons
B) 76.518 newtons
C) 111.834 newtons
Explanation:
A)
, where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:
![F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}](https://tex.z-dn.net/?f=F_g%3D%5Cdfrac%7B%286.67408%20%5Ctimes%2010%5E%7B-11%7D%29%287.8%29%2811.4%29%7D%7B%280.8%29%5E2%7D%5Capprox%209.273%20%5Ctimes%2010%5E%7B-9%7D)
B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.
C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.
Hope this helps!
Answer:
a) ![\frac{Ws}{Es} = \frac{200}{1+1.2s}](https://tex.z-dn.net/?f=%5Cfrac%7BWs%7D%7BEs%7D%20%20%3D%20%5Cfrac%7B200%7D%7B1%2B1.2s%7D)
b) attached below
c) type zero system
d) k > ![\frac{g}{200}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B200%7D)
e) The gain K increases above % error as the steady state speed increases
Explanation:
Given data:
Motor voltage = 12 v
steady state speed = 200 rad/s
time taken to reach 63.2% = 1.2 seconds
<u>a) The transfer function of the motor from voltage to speed</u>
let ;
be the transfer function of a motor
when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec
hence the transfer function of the motor from voltage to speed
= ![\frac{Ws}{Es} = \frac{200}{1+1.2s}](https://tex.z-dn.net/?f=%5Cfrac%7BWs%7D%7BEs%7D%20%20%3D%20%5Cfrac%7B200%7D%7B1%2B1.2s%7D)
<u>b) draw the block diagram of the system with plant controller and the feedback path </u>
attached below is the remaining part of the detailed solution
c) The system is a type-zero system because the pole at the origin is zero
d) ) k > ![\frac{g}{200}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B200%7D)
Answer:
Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.
Answer:
(a) dynamic viscosity = ![1.812\times 10^{-5}Pa-sec](https://tex.z-dn.net/?f=1.812%5Ctimes%2010%5E%7B-5%7DPa-sec)
(b) kinematic viscosity = ![1.4732\times 10^{-5}m^2/sec](https://tex.z-dn.net/?f=1.4732%5Ctimes%2010%5E%7B-5%7Dm%5E2%2Fsec)
Explanation:
We have given temperature T = 288.15 K
Density ![d=1.23kg/m^3](https://tex.z-dn.net/?f=d%3D1.23kg%2Fm%5E3)
According to Sutherland's Formula dynamic viscosity is given by
, here
μ = dynamic viscosity in (Pa·s) at input temperature T,
= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
= reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
![\mu _0=4\pi \times 10^{-7}](https://tex.z-dn.net/?f=%5Cmu%20_0%3D4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D)
= 291.15
when T = 288.15 K
For kinematic viscosity :
![\nu = \frac {\mu} {\rho}](https://tex.z-dn.net/?f=%5Cnu%20%3D%20%5Cfrac%20%7B%5Cmu%7D%20%7B%5Crho%7D)
![kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec](https://tex.z-dn.net/?f=kinemic%5C%20viscosity%3D%5Cfrac%7B1.812%5Ctimes%2010%5E%7B-5%7D%7D%7B1.23%7D%3D1.4732%5Ctimes%2010%5E%7B-5%7Dm%5E2%2Fsec)