Moment of inertia for one rod is expressed as:
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I(1)= I(end) + md^2 = (1\12)mL^2+m(L/2√2)^2=(5/24)mL^2
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Therefore, for two rods:
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I2 = 2I1 = (5/12)mL^2
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For the moment of inertia at the pivot point,
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I=I2+2md^2=(5/12)mL^2+2(m(L/2√2)^2)=(2/3)mL^2
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Substituting the equations above to the equation for frequency:
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f=(1/2π)√(2mgd/I)=(1/4π)√(6g/√2L)</span>
The equator would be warmer than the poles. :)
Acceleration = (change in speed) / (time for the change)
Change in speed = (speed after the change) - (speed before the change)
Change in speed = (65 m/s) - (35 m/s) = 30 m/s
Acceleration = (30 m/s) / (5 s)
<em>Acceleration = 6 m/s²</em>
Answer:
F = 1250000 N
Explanation:
If you have 50,000 S, by 25 M. It would be calculated to 1,250,000 N.
Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s