The old electricity meters with disk using sapphire bearings for friction to be minimal to not aggravate metering
Answer:
<em> think 2 also if not im so sorry but i think it is :)</em>
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L) (1)
200 = k(60 - L) (2)
(2)/(1): 2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
