<span>Negatively charged particles will go toward the positive end of the magnetic field and positively charged particles will go toward the negative end... for two reasons, because the positive is both repelled by the positive part of the field, and its also attracted by the negative end. oh, and on diagrams, the arrows of magnetic flux usually go from negative to positive.
Hope this helps(:</span>
Answer:
Explanation:
Given that,
Radius of the wheel, r = 20 cm = 0.2 m
Initial speed of the wheel, 
Displacement, 
To find,
The angular acceleration and the distance covered by the car.
Solution,
Let 
is the angular acceleration of the car. Using equation of rotational kinematics as :
Let t is the time taken by the car before coming to rest.
t = 30.39 seconds
Let v is the linear velocity of the car. So,
v = 150.79 m/s
Let d is the distance covered by the car. It can be calculated as :
d = 4582.5 meters
or
d = 4.58 km
Answer:
E = 3544.44 N/C
Explanation:
Given:
- charge Q = 2.2 *10^-6 C
- Length L = 1.3 m
Find:
The Electric Field strength E @ a = 1.8 m
Solution:
- The differential electric field dE due to infinitesimal charge dq can be considered as a point charge at a distance of r is given by:
dE = k*dq / r^2
- The charge Q is spread over entire length L, hence:
dq = (Q / L ) * dx
-The resulting dE:
dE = (k*Q/L)*(dx / r^2)
- point P lies on the x- axis with distance (x+a) from differential charge from:
dE = (k*Q/L)*(dx / (x+a)^2)
- Integrate dE over length 0 to L
E = (-k*Q/L)*( 1 / (x+a) )
E = (-k*Q/L)* (1 / a - 1 / (L+a))
E = (-k*Q/L)* (L / a(L+a))
E = (k*Q / a(L+a))
- Evaluate E @ a = 1.8 m
E =(8.99*10^9 * 2.2*10^-6 / 1.8*(1.3+1.8))
E = 3544.44 N/C
My guess would be c.africa
