Answer:
<h2>135,000 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 900 × 150
We have the final answer as
<h3>135,000 J</h3>
Hope this helps you
Answer:
x = 0.974 L
Explanation:
given,
length of inclination of log = 30°
mass of log = 200 Kg
rock is located at = 0.6 L
L is the length of the log
mass of engineer = 53.5 Kg
let x be the distance from left at which log is horizontal.
For log to be horizontal system should be in equilibrium
∑ M = 0
mass of the log will be concentrated at the center
distance of rock from CM of log = 0.1 L
now,
∑ M = 0



x = 0.974 L
hence, distance of the engineer from the left side is equal to x = 0.974 L
Answer:
Answer is a wave increasing in energy as it hits another wave.
Explanation:
I hope it's helpful!
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)
Acceleration occurs whenever the forces on an object are unbalanced.
It's the group of forces on the object that's either balanced or unbalanced.
There's no such thing as "an unbalanced force".