Answer:
current in series is 2.50 mA
current in parallel is 13.51 mA
Explanation:
given data
voltage = 5 V
resistors R1 = 1.5 kilo ohms
resistors R2 = 0.5 kilo ohms
to given data
current flow
solution
current flow in series is express as here
current = voltage / resistor .................1
put here all value in equation 1
current = 5 / (1.5 + 0.5)
current = 5 / 2.0
so current = 2.50 mA
and
current flow in parallel is express as
current = voltage / resistor ....................2
put here all value in equation 2
current = 5 / (1/ (1/1.5 + 1/0.5))
current = 5 / 0.37
so current = 13.31 mA
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Answer:
a) For y = 102 mA, R = 98.039 ohms
For y = 97 mA, R = 103.09 ohms
b) Check explanatios for b
Explanation:
Applied voltage, V = 10 V
For the first measurement, current 
According to ohm's law, V = IR
R = V/I
Here, 

For the second measurement, current 


b) ![y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy_%7B1%7D%20%26y_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D%20%5E%7BT%7D)
![y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy_%7B1%7D%20%5C%5Cy_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D)
![y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-3%7D%20%5C%5C97%2A10%5E%7B-3%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
A linear equation is of the form y = Gx
The nominal value of the resistance = 100 ohms
![x = \left[\begin{array}{ccc}100\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D100%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-3%7D%20%5C%5C97%2A10%5E%7B-3%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DG_%7B1%7D%20%5C%5CG_%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D100%5Cend%7Barray%7D%5Cright%5D%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DG_%7B1%7D%20%5C%5CG_%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-5%7D%20%5C%5C97%2A10%5E%7B-5%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
Answer:
160.75 N
Explanation:
The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.
As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.
When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:
F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N