Answer:
-30 N/C
Explanation:
Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m
Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V
Since E = -ΔV/Δx
substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -0.30 V/0.01 m
E = -30 V/m
Since 1 V/m = 1 N/C.
E = -30 N/C
So, the average electric field is -30 N/C
Answer:
180.4 m
Explanation:
The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:
h = 100 m
x =?
Height formula h:
Time t is cleared:
t = 4.51 sec
Horizontal distance formula x:
x = 40 m / sec x 4.51 sec
x = 180.4 m
Answer:
1.70 J
Explanation:
The heat dissipated is the difference in the kinetic energies.
This is given by
and are the initial and final velocities.
With <em>m</em> = 0.175 kg,
The negative sign appears because energy is lost.