They all share the way that they are fundamentally designed: if they are quite complex, they will share the same basic logic foundations, like the way that the programming languages work. They also all share the method of construction and common and fundamental electronic components, like resistors, capacitors and transistors. As we humans design them, they make logical sense to at least someone, and probably only discounting the internet, you can probably draw logic diagrams and whatever to represent how they work.
Because they are designed by Humans, in a way they all mimic how our brains and society work. Also, as yet there are no truly intelligent technological systems, and are only able to react to a situation how they have been programmed to do so.
Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
![Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%5Climits%5E2_0%20%7Bi%28t%29%7D%20%5C%2C%20dt%20%3D%20%5Cint%5Climits%5E2_0%20%7B6%2Ae%5E%28-2t%29%7D%20%5C%2C%20dt%20%3D%206%2A%5Cint%5Climits%5E2_0%20%7Be%5E%28-2t%29%7D%20%5C%2C%20dt)
- Integrate and evaluate the on the interval:
![= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C](https://tex.z-dn.net/?f=%3D%206%20%2A%20%28-0.5%29%2Ae%5E-2t%20%3D%20-%203%2A%28%201%20%2F%20e%5E4%20-%201%29%20%3D%202.945%20C)
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
![W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5E3_0%20%7BP%28t%29%7D%20%5C%2C%20dt%20%3D%20%5Cint%5Climits%5E2_0%20%7B-720%2Ae%5E%28-4t%29%7D%20%5C%2C%20dt%20%3D%20-720%2A%5Cint%5Climits%5E2_0%20%7Be%5E%28-4t%29%7D%20%5C%2C%20dt)
- Integrate and evaluate the on the interval:
![W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ](https://tex.z-dn.net/?f=W%20%3D%20-180%2Ae%5E-4t%20%3D%20-%20180%2A%28%201%20%2F%20e%5E12%20-%201%29%20%3D%20-180uJ)
Maybe it’s a vending machine, I’m confused too
Answer:
43248 newtons.
Explanation:
Force = mass x accelerations and units of force are newtons which are given in the question.
here mass = 125 of air and 2.2 of fuel, total = 125+2.2=127.5kg/s and the velocity of the exhaust is 340m/s.
force = 340m/s * 127.5kg/s = 43248 newtons technically this is wrong (observe units) but i will expalin how i have taken acceleration as a velocity here and mass/unit time as simply mass.
see force is mass times acceleration or deceleration, here our velocity is not changing therefore it is constant 340m/s but if it were to change and become 0 in one second then there would be -340m/s^2 (note the units ) of deceleration and there would be force associated with it and that force is what i have calculated here. similarly there would be mass in flow rate of mass per second, which is also in that one second of time.
let's calculate error.
error = (actual-calculated)/actual. = (43248-60000)/43248= -38.734% less is ofcourse greater than 2%.
So the load cell is not reading correct to within 2% and it should read 43248newtons.