Explanation:
Cell is a structural and fundamental unit mass of the body
The study of EM is essential to understanding the properties of light, its propagation through tissue, scattering and absorption effects, and changes in the state of polarization. ... Since light travels much faster than sound, detection of the reflected EM radiation is performed with interferometry.
Answer:
The electric flux is ![280\ \rm N.m^2/C](https://tex.z-dn.net/?f=280%5C%20%5Crm%20N.m%5E2%2FC)
Explanation:
Given:
- Radius of the disc R=0.50 m
- Angle made by disk with the horizontal
![\theta=30^\circ](https://tex.z-dn.net/?f=%5Ctheta%3D30%5E%5Ccirc)
- Magnitude of the electric Field
![E=713.0\ \rm N/C](https://tex.z-dn.net/?f=E%3D713.0%5C%20%5Crm%20N%2FC)
The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows
![\phi=\int E.dA](https://tex.z-dn.net/?f=%5Cphi%3D%5Cint%20E.dA%20)
where
is the total Electric Flux- E is the Electric Field
- dA is the Area through which the electric flux is to be calculated.
Now according to question we have
![=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C](https://tex.z-dn.net/?f=%3DEA%5Ccos%5Ctheta%20%5C%5C%3D713%5Ctimes%203.14%5Ctimes%200.5%5E2%20%5Ctimes%20%5Ccos60%5E%5Ccirc%5C%5C%3D280%5C%20%5Crm%20N.m%5E2%2FC)
Hence the electric flux is calculated.
Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°
Answer:
(a) θ = 33.86°
(b) Ay = 49.92 N
Explanation:
You have that the magnitude of a vector is A = 89.6 N
The x component of such a vector is Ax = 74.4 N
(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:
(1)
Ax: x component of vector A
A: magnitude of vector A
θ: angle between vector A and the x axis
You solve the equation (1) for θ, by using the inverse of cosine function:
![\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%28%5Cfrac%7BA_x%7D%7BA%7D%29%3Dcos%7B-1%7D%28%5Cfrac%7B74.4N%7D%7B89.6N%7D%29%5C%5C%5C%5C%5Ctheta%3D33.86%5C%C2%B0)
the angle between the A vector and the x axis is 33.86°
(b) The y component of the vector is given by:
![A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N](https://tex.z-dn.net/?f=A_y%3DAsin%5Ctheta%5C%5C%5C%5CA_y%3D%2889.6N%29sin%2833.86%5C%C2%B0%29%3D49.92N)
the y comonent of the vecor is Ay = 49.92 N