This question involves the concept of the scalar product.
The magnitude of the scalar product will be "0".
<h3>SCALAR PRODUCT</h3>
The scalar product, also known as the dot product of the two vectors is given by the following formula:
![A.B = |A||B|Cos\theta](https://tex.z-dn.net/?f=A.B%20%3D%20%7CA%7C%7CB%7CCos%5Ctheta)
where,
- A.B = Scalar product = ?
- |A| = Mangnitude of vector A = 1 unit
- |B| = Magnitude of Vector B = 1 unit
- θ = Angle between vectors = 90°
Therefore,
![A.B = (1)(1)Cos90^o = (1)(1)(0)](https://tex.z-dn.net/?f=A.B%20%3D%20%281%29%281%29Cos90%5Eo%20%3D%20%281%29%281%29%280%29)
A.B = 0
Learn more about scalar product here:
brainly.com/question/6849226
Answer:
The angular acceleration of the wheel is 15.21 rad/s².
Explanation:
Given that,
Time = 5 sec
Final angular velocity = 96.0 rad/s
Angular displacement = 28.0 rev = 175.84 rad
Let
be the angular acceleration
We need to calculate the angular acceleration
Using equation of motion
![\theta=\omega_{i} t+\dfrac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_%7Bi%7D%20t%2B%5Cdfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
Put the value in the equation
![175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2](https://tex.z-dn.net/?f=175.84%3D%5Comega_%7Bi%7D%5Ctimes%205%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%5Calpha%5Ctimes%285%29%5E2)
......(I)
Again using equation of motion
![\omega_{f}=\omega_{i}+\alpha t](https://tex.z-dn.net/?f=%5Comega_%7Bf%7D%3D%5Comega_%7Bi%7D%2B%5Calpha%20t)
Put the value in the equation
![96.0=\omega_{i}+\alpha \times 5](https://tex.z-dn.net/?f=96.0%3D%5Comega_%7Bi%7D%2B%5Calpha%20%5Ctimes%205)
On multiply by 5 in both sides
....(II)
On subtract equation (I) from equation (II)
![480-175.84=\alpha(25-5)](https://tex.z-dn.net/?f=480-175.84%3D%5Calpha%2825-5%29)
![304.16=\alpha\times20](https://tex.z-dn.net/?f=304.16%3D%5Calpha%5Ctimes20)
![\alpha=\dfrac{304.16}{20}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B304.16%7D%7B20%7D)
![\alpha=15.21\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%3D15.21%5C%20rad%2Fs%5E2)
Hence, The angular acceleration of the wheel is 15.21 rad/s².
Answer:
b. 1.1 m
Explanation:
It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,
![s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)](https://tex.z-dn.net/?f=s_%7B1%7D%20%2B%20s_%7B2%7D%20%3D%203%5C%20m%5C%5C%5C%5Cs_%7B2%7D%20%3D%203%5C%20m%20-%20s_%7B1%7D%5C%20---------%20eqn%281%29)
where,
s₁ = distance of fulcrum from left mass
s₂ = distance of fulcrum from right mass
In order to achieve balance, the torque due to both masses must be equal:
![T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%20T_%7B2%7D%5C%5Cm_%7B1%7Ds_%7B1%7D%20%3D%20m_%7B2%7Ds_%7B2%7D%5C%5C%2825%5C%20kg%29%28s_%7B1%7D%29%20%3D%20%2815%5C%20kg%29%28s_%7B2%7D%29%5C%5C%5C%5C%5Cfrac%7B15%5C%20kg%7D%7B25%5C%20kg%7D%28s_%7B2%7D%29%20%3D%20s_%7B1%7D%5C%5C%5C%5Cusing%5C%20eqn%281%29%3A%5C%5C%280.6%29%283%5C%20m%20-%20s_%7B1%7D%29%20%3D%20s_%7B1%7D%5C%5C1.8%5C%20m%20%3D%201.6%5C%20s_%7B1%7D%5C%5Cs_%7B1%7D%20%3D%20%5Cfrac%7B1.8%5C%20m%7D%7B1.6%7D)
s₁ = 1.1 m
Hence, the correct option is:
<u>b. 1.1 m</u>
i think the data is not complete but that's according to me