Answer:
B. The more optically dense the material, the higher the refractive index.
Explanation:
Answer:
A. Mrŵ² = ųMg
Ŵ = (ųg/r)^½
B.
Ŵ =[ (g /r)* tan á]^½
Explanation:
T.v.= centrepetal force = mrŵ²
Where m = mass of block,
r = radius
Ŵ = angular momentum
On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.
So there for
Mrŵ² = ųMg
Ŵ = (ųg/r)^½
g = Gravitational pull
ų = coefficient of friction.
B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible
So therefore
N *(sin á) = mrŵ² .....equ 1
Since the car does not slide the net vertical forces must be equal and opposite so therefore
N*(cos á) = mg.....equ 2
Where N is the reaction force of the car on the surface.
Equ 2 becomes N = mg/cos á
Substituting N into equation 1
mg*(sin á /cos á) =mrŵ²
Tan á = rŵ²/g
Ŵ =[ (g /r)* tan á]^½
Answer:
The magnitude of the EMF is 0.00055 volts
Explanation:
The induced EMF is proportional to the change in magnetic flux based on Faraday's law:
![emf\,=-\,N\, \frac{d\Phi}{dt}](https://tex.z-dn.net/?f=emf%5C%2C%3D-%5C%2CN%5C%2C%20%5Cfrac%7Bd%5CPhi%7D%7Bdt%7D)
Since in our case there is only one loop of wire, then N=1 and we get:
![emf\,=-\,N\, \frac{d\Phi}{dt}](https://tex.z-dn.net/?f=emf%5C%2C%3D-%5C%2CN%5C%2C%20%5Cfrac%7Bd%5CPhi%7D%7Bdt%7D)
We need to express the magnetic flux given the geometry of the problem;
where A is the area of the coil that remains unchanged with time, and B is the magnetic field that does change with time. Therefore the equation for the EMF becomes:
![emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts](https://tex.z-dn.net/?f=emf%5C%2C%3D-%5C%2CN%5C%2C%20%5Cfrac%7Bd%5CPhi%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%5CPhi%7D%7Bdt%7D%20%3D-%5Cfrac%7Bd%5C%2C%28B%5C%2CA%29%7D%7Bdt%7D%20%3D-%5C%2CA%5C%2C%5Cfrac%7Bd%5C%2C%28B%29%7D%7Bdt%7D%3D-%201%5C%2Cm%5E2%282%5C%2C%5C%2CT%2Fh%7D%29%3D%20-2%5C%2C%5C%2Cm%5E2%5C%2CT%2F%283600%5C%2C%5C%2Cs%29%3D%20-0.00055%5C%2CVolts)
To answer the question the antenna would probably burn up in the atmosphere. I have no idea how to get that thing on Earth if it's by itself. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.