The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
where
is the angle between the directions of v and B.
1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is
. In this case, v and B are perpendicular, so
, therefore we have:
2) In this second case, the angle between v and B is
. The charge is now
, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
PE= 3kg x 10N/kg x 10m
= 300J
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
Answer:
1/8 x C
Explanation:
The capacitance of parallel plate capacitor
= ε₀ A /d where A is area of plate and d is distance between plate.
for capacitor 1
C = ε₀ A /d
For capacitor 2
radius = R/2
Area = A / 4
Capacitance
= ε₀ (A/4) x ( 1 / 2d )
= ( 1 / 8) x (ε₀ A /d)
= 1/8 x C
