We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2
Answer:
Explanation:
We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.
Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C
Answer:
the active region is bound by cutoff region and saturation or power dissipation region.
Explanation:
Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a
h=5.15 m
The depth is 5.15 m.
