Fulcrum need to be positioned balanced with weight on both the sides following law of lever.
What is the physical law of the lever?
- It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
- The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
- The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
- Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.
Learn more about the Fulcrum with the help of the given link:
brainly.com/question/16422662
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Explanation:
It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.
The charge per unit length of the wire is 
and the net charge per unit length is 
.
We know that there exist zero electric field inside the metal cylinder.
(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let 
are the charge per unit length on the inner and outer surfaces of the cylinder.
For inner surface,
For outer surface,
(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :
Hence, this is the required solution.
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
The answer is B (The second one). I'm not sure though.
Answer:
F = 1.047 10⁻² N
Explanation:
Let's use kinematics to find the angular acceleration
w = w₀ + α t
as for rest w₀ = 0
w = α t
α = w / t
let's reduce the magnitudes to the SI system
w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s
m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg
r = 10.0 cm (1 m / 100 cm) = 0.100 m
let's calculate
α = 104.72 / 1
α = 104.72 rad / s²
angular and linear variables are related
a = α r
a = 104.72 0.100
a = 10.47 m / s²
finally we substitute in Newton's second law
F = 1 10⁻³ 10.47
F = 1.047 10⁻² N
