Answer:
<em> -18896.49 V/m</em>
<em></em>
Explanation:
Distance between the two plates = 10 cm = 10 x m = 0.1 m
Also, one of the plates is taken as<em> zero volt.</em>
a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V
b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V
<em>Potential field strength = -dV/dx</em>
where dV is voltage difference between these points,
dx is the difference in distance between these points
For the first case above,
potential field strength = -393/0.0705 = -5574.46 V/m
For the second case ,
potential field strength = -393/0.0295 = -13322.03 V/m
Magnitude of the field strength across the plates will be
-5574.46 + (-13322.03) = -5574.46 + 13322.03 =<em> -18896.49 V/m</em>
Answer:
<em>a = 7.6\ mph/s</em>
Explanation:
<u>Motion With Constant Acceleration </u>
It's a type of motion in which the velocity of an object changes uniformly in time.
The equation that describes the change of velocities is:
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
Solving the equation [for a:
The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:
a = 7.6\ mph/s
Answer:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
Formula:
As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.