Answer:
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Explanation:
Answer:
T = 1.1285 10⁻² day
Explanation:
For this exercise the forces in the premiere are internal, so the angular momentum is conserved
L₀ = I₀ w₀
L = I w
L₀ = L
I₀ w₀ = I w
Angular velocity and period are related
w₀ = 2π / T₀
w = 2π / T
The moment of inertia of a sphere is
I₀ = 2/5 M R²
I = 2/5 m r²
If we assume that the mass of the star does not change in the transformation
We substitute
2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T
R² /T₀ = r² / T
T = (r / R)² T₀
T = (6.1 / 2.0 104) 37
T = 1.1285 10⁻² day
Answer:
a. 5.23 m/s² b. 44.23 N
Explanation:
a. What is the centripetal acceleration of the hammer?
The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.
Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.
So a = rω²
= 1.4 m × (3.74 rad/s)²
= 5.23 m/s²
b. What is the tension in the chain?
The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.
So, T = 8.45 kg × 5.23 m/s²
= 44.23 N
Answer:
Explanation:
Given data:
v = 220 rms
power factor = 0.65
P = 1250 W
New power factor is 0.9 lag
we knwo that
s = 1923.09 < 49.65^o
s = [1250 + 1461 j] vA
![P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]](https://tex.z-dn.net/?f=P.F%20new%20%3D%20cos%20%5Btan%5E%7B-1%7D%20%5Cfrac%7BQ_%7Bnew%7D%7D%7BP%7D%5D)
solving for 
![Q_{new} = P tan [cos^{-1} P.F new]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%20P%20tan%20%5Bcos%5E%7B-1%7D%20P.F%20new%5D)
![Q_{new} = 1250 [tan[cos^{-1}0.9]]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%201250%20%5Btan%5Bcos%5E%7B-1%7D0.9%5D%5D)
Faraday
