Answer:
83.3 Wb
Explanation:
The magnetic flux linkage through the coil is given by:
![N\phi = BAN sin \theta](https://tex.z-dn.net/?f=N%5Cphi%20%3D%20BAN%20sin%20%5Ctheta)
where
B is the magnetic field strength
A is the cross sectional area
N is the number of turns in the coil
is the angle between the direction of the field and the normal to the coil
In this problem:
B = 1.74 T
A = 0.133 m^2
N = 360
![\theta=90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7B%5Ccirc%7D)
Therefore, the magnetic flux linkage is
![N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb](https://tex.z-dn.net/?f=N%5Cphi%20%3D%20%281.74%20T%29%280.133%20m%5E2%29%28360%29%20sin%2090%5E%7B%5Ccirc%7D%3D83.3%20Wb)
Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)
Formula (Kinetic energy)
![E_{k} = \frac{m*s^2}{2}](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%20%20%5Cfrac%7Bm%2As%5E2%7D%7B2%7D%20)
Solving:
![E_{k} = \frac{m*s^2}{2}](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%20%5Cfrac%7Bm%2As%5E2%7D%7B2%7D%20)
![E_{k} = \frac{1*3^2}{2}](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%20%20%5Cfrac%7B1%2A3%5E2%7D%7B2%7D%20)
![E_{k} = \frac{1*9}{2}](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%20%20%5Cfrac%7B1%2A9%7D%7B2%7D%20)
![E_{k} = \frac{9}{2}](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%20%20%5Cfrac%7B9%7D%7B2%7D%20)
Answer:
because it's north
Explanation:
u if yyggggggggggggggg hgfth
Answer: The answer is D: 300,000km/s
Explanation: