An eclipse is a phenomenon of an astronomical object being obscured by something. In this case, it is the Moon that is being obscured from sunlight by Earth's shadow. Answer is D.
Answer:
c) true. With the magnetic field you can determine the Poynting vector and this is equal to the intensity of the wave
Explanation:
An electromagnetic wave carries energy that is given by the vector of poyntíng
S = 1 / μ₀ E x B
The electric and magnetic fields are related
E = c B
Substituting
S = c /μ₀ B² / 2
The two comes from the average value
Let's examine the claims
a) false to determine the wavelength the frequency is needed
b) False. To find the electric field you need the magnetic field
c) true. With the magnetic field you can determine the Poynting vector and this is equal to the intensity of the wave
Answer:
Part a)
![t_1 = \frac{\omega_1}{\alpha}](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cfrac%7B%5Comega_1%7D%7B%5Calpha%7D)
Part b)
![\theta = \frac{1}{2}\alpha t_1^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t_1%5E2)
Part c)
![t = \frac{t_1}{5}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bt_1%7D%7B5%7D)
Explanation:
Part a)
As we know that it is having constant torque so here the time taken by it to accelerate is given as
![\omega_f = \omega_i + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_i%20%2B%20%5Calpha%20t)
![\omega_1 = 0 + \alpha t_1](https://tex.z-dn.net/?f=%5Comega_1%20%3D%200%20%2B%20%5Calpha%20t_1)
![t_1 = \frac{\omega_1}{\alpha}](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cfrac%7B%5Comega_1%7D%7B%5Calpha%7D)
Part b)
angular displacement is given as
![\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega_i%20t_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Calpha%29%20t_1%5E2)
![\theta = 0 + \frac{1}{2}\alpha t_1^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t_1%5E2)
![\theta = \frac{1}{2}\alpha t_1^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t_1%5E2)
Part c)
As we know that the angular deceleration produced by the brakes is given as
![\alpha_d = - 5\alpha](https://tex.z-dn.net/?f=%5Calpha_d%20%3D%20-%205%5Calpha)
now we have
![\omega_f = \omega _i + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega%20_i%20%2B%20%5Calpha%20t)
![0 = \omega_1 - 5 \alpha_1 t](https://tex.z-dn.net/?f=0%20%3D%20%5Comega_1%20-%205%20%5Calpha_1%20t)
![t = \frac{\omega_1}{5 \alpha}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Comega_1%7D%7B5%20%5Calpha%7D)
As we know that
![t_1 = \frac{\omega_1}{\alpha}](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cfrac%7B%5Comega_1%7D%7B%5Calpha%7D)
so we have
![t = \frac{t_1}{5}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bt_1%7D%7B5%7D)
Answer:
the two beams must travel paths that differ by one-half of a wavelength.