Acharged particle moves at 2.5 × 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10–2 t. if the magnetic force is 7.5 × 10–2 n, what is the magnitude of the charge?

the magnitude of charge=q=8.76 x 10⁻⁵C

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge

v= velocity= 2.5 x 10⁴ m/s

B= magnetic field strength= 8.1 x 10⁻²T

Fm= magnetic force= 7.5 x 10⁻² N

θ=25°

so 7.5 x 10⁻² =q (2.5 x 10⁴ ) (8.1 x 10⁻²) sin25

q=8.76 x 10⁻⁵C

answer is C on edg

Explanation:

the most reliable source of information.is wikipedia

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14