Answer:
because fm radio waves are affected by tall buildings or tall objects but am radio wave are affected storms
Explanation:
Answer:
![\beta=B=8.05\mu T](https://tex.z-dn.net/?f=%5Cbeta%3DB%3D8.05%5Cmu%20T)
Explanation:
The density of the magnetic flux is given by the following formula:
![\beta=\frac{\Phi_B}{A}=\frac{ABcos\theta}{A}=Bcos\theta](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cfrac%7B%5CPhi_B%7D%7BA%7D%3D%5Cfrac%7BABcos%5Ctheta%7D%7BA%7D%3DBcos%5Ctheta)
The normal vector A and the vector of the magnitude of the magnetic field are perpendicular, then, the angle is zero:
The magnitude of the magnetic field is calculated by using the formula for B at a distance of x to a point in the plane of the loop:
![B=\frac{\mu_oIR^2}{2(x^2+R^2)^{3/2}}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cmu_oIR%5E2%7D%7B2%28x%5E2%2BR%5E2%29%5E%7B3%2F2%7D%7D)
For x = 0 you have:
![B=\frac{\mu_oIR^2}{2R^3}=\frac{\mu_oI}{2R}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cmu_oIR%5E2%7D%7B2R%5E3%7D%3D%5Cfrac%7B%5Cmu_oI%7D%7B2R%7D)
R is the radius of the circular loop and its values is:
![R=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{0.5m^2}{\pi}}=0.39m](https://tex.z-dn.net/?f=R%3D%5Csqrt%7B%5Cfrac%7BA%7D%7B%5Cpi%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.5m%5E2%7D%7B%5Cpi%7D%7D%3D0.39m)
Then, you replace in the equation for B with mu_o = 4\pi*10^-7 T/A:
![B=\frac{(4\pi*10^{-7}T/A)(5A)}{2(0.39m)}=8.05*10^{-6}T=8.05\mu T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%285A%29%7D%7B2%280.39m%29%7D%3D8.05%2A10%5E%7B-6%7DT%3D8.05%5Cmu%20T)
and the density of the magnetic flux is
![\beta=B=8.05\mu T](https://tex.z-dn.net/?f=%5Cbeta%3DB%3D8.05%5Cmu%20T)
Answer:
= 625 nm
Explanation:
We now that for
for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....
d= distance between the slits, λ= wavelength of incident ray
for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.
Given
d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m
applying formula
λ= (d*y)/(D*n)
putting values we get
![\lambda = \frac{1.19\times10^{-3}\times4.97\times10^{-2}}{9.47\times10}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B1.19%5Ctimes10%5E%7B-3%7D%5Ctimes4.97%5Ctimes10%5E%7B-2%7D%7D%7B9.47%5Ctimes10%7D)
on solving we get
= 625 nm
R=10
F=R/2
F=10/2=5
F=-5(CONCAVE MIRROR)
U=-8(CONCAVE MIRROR)
HEIGHT OF OBJECT=1.5
V=?
HEIGHT OF IMAGE=?
I/F=1/U+1/V
-I/5=-1/8-1/V
-1/V=-1/5+1/8
-1/V=-8+5/40
-1/V=-3/40
1/V=3/40
V=40/3
HEIGHT OF IMAGE/HEIGHT OF OBJECT =-V/U
HEIGHT OF IMAGE=40/3*1/-8*15/10
=-20/8
=-2.5
The answer is C due to the kinetic movement of the rc car driving around and the electrochemical energy turning into electrical energy.