Answer:
Slick and smooth surfaces mediate friction to a low, allowing an object to speed across it with low resistance.
Explanation:
Compute the components of the given vectors. Let
denote the plane's velocity vector, and
the wind. Then
![P=\langle P_x,P_y\rangle=\langle250\cos(-40^\circ),250\sin(-40^\circ)\rangle](https://tex.z-dn.net/?f=P%3D%5Clangle%20P_x%2CP_y%5Crangle%3D%5Clangle250%5Ccos%28-40%5E%5Ccirc%29%2C250%5Csin%28-40%5E%5Ccirc%29%5Crangle)
![\implies P=\langle191.5,-160.7\rangle](https://tex.z-dn.net/?f=%5Cimplies%20P%3D%5Clangle191.5%2C-160.7%5Crangle)
![W=\langle35\cos(-120^\circ),35\sin(-120^\circ)\rangle=\langle-17.5,-30.3\rangle](https://tex.z-dn.net/?f=W%3D%5Clangle35%5Ccos%28-120%5E%5Ccirc%29%2C35%5Csin%28-120%5E%5Ccirc%29%5Crangle%3D%5Clangle-17.5%2C-30.3%5Crangle)
The resultant velocity (rounded) is
![P+W=\langle174,-191\rangle](https://tex.z-dn.net/?f=P%2BW%3D%5Clangle174%2C-191%5Crangle)
with magnitude
and direction
, or about 258 mi/hr at 47.7 degrees south of east.
Answer:
true , I searched and got u the answer
The answer to your question is a 25.6 HZ
Explanation:
It is given that,
Diameter of the circular loop, d = 1.5 cm
Radius of the circular loop, r = 0.0075 m
Magnetic field, ![B=2.7\ mT=2.7\times 10^{-3}\ T](https://tex.z-dn.net/?f=B%3D2.7%5C%20mT%3D2.7%5Ctimes%2010%5E%7B-3%7D%5C%20T)
(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :
![B=\dfrac{\mu_o I}{2r}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_o%20I%7D%7B2r%7D)
![I=\dfrac{2Br}{\mu_o}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2Br%7D%7B%5Cmu_o%7D)
![I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%5Ctimes%202.7%5Ctimes%2010%5E%7B-3%7D%5Ctimes%200.0075%7D%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%7D)
I = 32.22 A
(b) The magnetic field on a current carrying wire is given by :
![B=\dfrac{\mu_o I}{2\pi r}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%20r%7D)
![r=\dfrac{\mu_o I}{2\pi B}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%20B%7D)
![r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2032.22%7D%7B2%5Cpi%20%5Ctimes%202.7%5Ctimes%2010%5E%7B-3%7D%7D)
r = 0.00238 m
![r=2.38\times 10^{-3}\ m](https://tex.z-dn.net/?f=r%3D2.38%5Ctimes%2010%5E%7B-3%7D%5C%20m)
Hence, this is the required solution.