Answer:
perpendicular to
Explanation:
it means perpendicular to .....should u come across something like this / / , this one means parallel to .....
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
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:
53.13 °
Explanation:
In order to do this, we just need to apply the following:
tanα = Dy/Dx
Where:
Vy: speed of the ball in the y axis.
Vx: speed of the ball in the x axis.
At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.
According to this, then:
tanα = (40/30)
tanα = 1.3333
α = tan⁻¹(1.3333)
<h2>
α = 53.13°</h2>
This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.
Hope this helps