here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.
So angle must be more than 270 degree and less than 360 degree
Now in order to find the value we can say that
![tan\theta = \frac{opposite\: side}{adjacent\: side}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7Bopposite%5C%3A%20side%7D%7Badjacent%5C%3A%20side%7D)
![tan\theta = \frac{8.6}{6.1}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7B8.6%7D%7B6.1%7D)
![\theta = tan^{-1}1.41](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D1.41)
![\theta = 54.65^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2054.65%5E0)
so it is inclined at above angle with X axis in fourth quadrant
Now if angle is to be measured counterclockwise then its magnitude will be
![\theta = 360 - 54.65 = 305.3^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20360%20-%2054.65%20%3D%20305.3%5E0)
so the correct answer will be 305 degree
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
In a direct current, the electric charge, or current, only flows in one direction. In an alternating current, the electric charge changes periodically.
Answer:
![E_y=1175510.2\ N.C^{-1}](https://tex.z-dn.net/?f=E_y%3D1175510.2%5C%20N.C%5E%7B-1%7D)
The Magnitude of electric field is in the upward direction as shown directly towards the charge
.
Explanation:
Given:
- side of a square,
![a=52.5\ cm](https://tex.z-dn.net/?f=a%3D52.5%5C%20cm)
- charge on one corner of the square,
![q_1=+45\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_1%3D%2B45%5Ctimes%2010%5E%7B-6%7D%5C%20C)
- charge on the remaining 3 corners of the square,
![q_2=q_3=q_4=-27\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_2%3Dq_3%3Dq_4%3D-27%5Ctimes%2010%5E%7B-6%7D%5C%20C)
<u>Distance of the center from each corners</u>![=\frac{1}{2} \times diagonals](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20diagonals)
![diagonal=\sqrt{52.5^2+52.5^2}](https://tex.z-dn.net/?f=diagonal%3D%5Csqrt%7B52.5%5E2%2B52.5%5E2%7D)
![diagonal=74.25\cm=0.7425\ m](https://tex.z-dn.net/?f=diagonal%3D74.25%5Ccm%3D0.7425%5C%20m)
∴Distance of center from corners, ![b=0.3712\ m](https://tex.z-dn.net/?f=b%3D0.3712%5C%20m)
Now, electric field due to charges is given as:
![E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Ctimes%20%5Cfrac%7Bq%7D%7Bb%5E2%7D)
<u>For charge
we have the field lines emerging out of the charge since it is positively charged:</u>
![E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}](https://tex.z-dn.net/?f=E_1%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cfrac%7B45%5Ctimes%2010%5E%7B-6%7D%7D%7B0.3712%5E2%7D)
<u>Force by each of the charges at the remaining corners:</u>
![E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}](https://tex.z-dn.net/?f=E_2%3DE_3%3DE_4%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cfrac%7B27%5Ctimes%2010%5E%7B-6%7D%7D%7B0.3712%5E2%7D)
<u> Now, net electric field in the vertical direction:</u>
![E_y=E_1-E_4](https://tex.z-dn.net/?f=E_y%3DE_1-E_4)
![E_y=1175510.2\ N.C^{-1}](https://tex.z-dn.net/?f=E_y%3D1175510.2%5C%20N.C%5E%7B-1%7D)
<u>Now, net electric field in the horizontal direction:</u>
![E_y=E_2-E_3](https://tex.z-dn.net/?f=E_y%3DE_2-E_3)
![E_y=0\ N.C^{-1}](https://tex.z-dn.net/?f=E_y%3D0%5C%20N.C%5E%7B-1%7D)
So the Magnitude of electric field is in the upward direction as shown directly towards the charge
.