In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω
Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂
The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁
Hence, the protostar will be rotating 16 </span><span>times faster.</span>
<u>Given that</u>
mass (m) = 1300 Kg ,
height (h) = 1500 m
Determine the potential energy ?
P.E = m × g × h
= 1300 × 9.81 × 1500
= 19129500 Joules
= 19129.5 KJ
Answer:
2.286 ohm
Explanation:
R1 = 16 ohm
R2 = 8 ohm
R3 = 4 ohm
They all are connected in parallel combination
Let the equivalent resistance is R.
1/R = 1/R1 + 1/R2 + 1/R3
1/R = 1/16 + 1/8 + 1/4
1/R = (1 + 2 + 4) / 16
1/R = 7 / 16
R = 16/7 = 2.286 ohm
Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s