Answer:
The ideal transformer has no resistance, but in the actual transformer, there is always some resistance to the primary and secondary windings. For making the calculation easy the resistance of the transformer can be transferred to the either side.
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>
![v_y(t)=vo*sin(A)-g*t](https://tex.z-dn.net/?f=v_y%28t%29%3Dvo%2Asin%28A%29-g%2At)
the velocity is Zero when the projectile reach in the maximum altitude:
![0=vo-gt\\t=\frac{vo}{g}](https://tex.z-dn.net/?f=0%3Dvo-gt%5C%5Ct%3D%5Cfrac%7Bvo%7D%7Bg%7D)
When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>
![d_x(t)=vo*cos(A)*t\\](https://tex.z-dn.net/?f=d_x%28t%29%3Dvo%2Acos%28A%29%2At%5C%5C)
R=Range
![R=d_x(t=2*\frac{vo}{g})](https://tex.z-dn.net/?f=R%3Dd_x%28t%3D2%2A%5Cfrac%7Bvo%7D%7Bg%7D%29)
![R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}](https://tex.z-dn.net/?f=R%3Dvo%2Acos%28A%29%2A2%5Cfrac%7Bvo%7D%7Bg%7D%20%5C%5C%5C%5CR%3D%5Cfrac%7B%28vo%29%5E%7B2%7D%2A2%2A%20sin%28A%29cos%28A%29%7D%7Bg%7D%20%5C%5C%5C%5CR%3D%5Cfrac%7B%28vo%29%5E%7B2%7D%20sin%282A%29%7D%7Bg%7D)
**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile
![\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\](https://tex.z-dn.net/?f=%5Cfrac%7Bvo%5E%7B2%7D%20%7D%7Bg%7D%20%3D%28%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%20%7D%29%20%5Cfrac%7Bvo%5E%7B2%7D%20%2Asin%282B%29%7D%7Bg%7D%5C%5C%5C%5C1%3D%20%5Cfrac%7B2%20%7D%7B%5Csqrt%7B3%7D%7D%20%2Asin%282B%29%5C%5C%5C%5Csin%282B%29%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%5C%5C%5C)
2B=60°
B=30°
745.699872 watts. There is no direct conversion to determine the amount of amps in 1 horsepower
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)