Answer:
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :
Here, 
is rate of increasing potential difference
So
So, the value of displacement current is 
.
m = mass of the penny
r = distance of the penny from the center of the turntable or axis of rotation
w = angular speed of rotation of turntable
F = centripetal force experienced by the penny
centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as
F = m r w²
in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .
hence greater the distance from center , greater will be the centripetal force to remain in place.
So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.
Answer:
a = 0.55 m / s²
Explanation:
The centripetal acceleration is given by the relation
a = v² / r
angular and linear velocities are related
v = w r
we substitute
a = w² r
In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system
w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s
let's calculate
a = 0.105² 50.0
a = 0.55 m / s²
A descriptive observation may very well be a mixture of both quantitative and qualitative as it can utilize elements of both types. Qualitative deals with the kinds of observations that cannot be measured in numerical form. Quantitative data is just that.
