(a) The amount of work required to change the rotational rate is 0.0112 J.
(b) The decrease in the rotational inertia when the outermost particle is removed is 64.29%.
<h3>
Moment of inertia of the rod</h3>
The moment of inertia of the rod from the axis of rotation is calculated as follows;
I = md² + m(2d)² + m(3d)²
where;
- m is mass = 10 g = 0.01 kg
- d = 3 equal division of the length
d = 6/3 = 2 cm = 0.02 m
I = md²(1 + 2² + 3²)
I = 14md²
I = 14(0.01)(0.02)²
I = 5.6 x 10⁻⁵ kg/m³
<h3>Work done to change the rotational rate</h3>
K.E = ¹/₂Iω²
K.E = ¹/₂(5.6 x 10⁻⁵)(60 - 40)²
K.E = ¹/₂(5.6 x 10⁻⁵)(20)²
K.E = 0.0112 J
<h3>Percentage decrease of rotational inertia when the outermost particle is removed</h3>
I₂ = md² + m(2d)²
I₂ = 5md²
ΔI = 14md² - 5md²
ΔI = 9md²
η = (ΔI/I) x 100%
η = (9md²/14md²) x 100%
η = 64.29 %
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Answer:
Explanation:
Using Hooke's law
Elastic potential energy = 1/2 K x²
K is elastic constant of the spring
x is the extension of the spring
a) The elastic potential energy when the spring is compressed twice as much Uel = 1/2 k (2x₀) ² = 4 (1/2 kx₀²)= 4 U₀
b) when is compressed half as much Uel = 1/2 k 
= 
( U₀)
c) make x₀ subject of the formula in the equation for elastic potential
x₀ =
x, the amount it will compressed to tore twice as much energy = 
x = √2 x₀
d) x₁, the new length it must be compressed to store half as much energy = 
x₁ = 
x₀
Convert cm into meters,
50/100=0.5m
work done= 10x0.5
=5J
we dont consider the weight of the spring as it acts downwards.
Answer:
I₁ = 1.6 A (through 7 Ohm Resistor)
I₂ = 1.3 A (through 8 Ohm Resistor)
I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)
Explanation:
Here we consider two loops doe applying Kirchhoff's Voltage Law (KVL). The 1st loop is the left side one with a voltage source of 12 V and the 2nd Loop is the right side one with a voltage source of 9 V. We name the sources and resistor's as follows:
R₁ = 7 Ω
R₂ = 4 Ω
R₃ = 8 Ω
V₁ = 12 V
V₂ = 9 V
Now, we apply KVL to 1st Loop:
V₁ = I₁R₁ + (I₁ - I₂)R₂
12 = 7I₁ + (I₁ - I₂)(4)
12 = 7I₁ + 4I₁ - 4I₂
I₁ = (12 + 4 I₂)/11 ------------ equation (1)
Now, we apply KVL to 2nd Loop:
V₂ = (I₂ - I₁)R₂ + I₂R₃
9 = (I₂ - I₁)(4) + 8I₂
9 = 4I₂ - 4I₁ + 8I₂
9 = 12I₂ - 4I₁ -------------- equation (2)
using equation (1)
9 = 12I₂ - 4[(12 + 4 I₂)/11]
99 = 132 I₂ - 48 - 16 I₂
147 = 116 I₂
I₂ = 147/116
I₂ = 1.3 A
use this value in equation 2:
9 = 12(1.3 A) - 4I₁
4I₁ = 15.6 - 9
I₁ = 6.6 A/4
I₁ = 1.6 A
Hence, the currents through all resistors are:
<u>I₁ = 1.6 A (through 7 Ohm Resistor)</u>
<u>I₂ = 1.3 A (through 8 Ohm Resistor)</u>
<u>I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)</u>
Answer: wavelength=velocity×period
Explanation:the relation between velocity, wavelength and period is
Wavelength=velocity×period
