Answer:
The kinetic energy lost in the collision is 48 J
Explanation:
Given;
mass of the first ball, m₁ = 2.0 kg
mass of the second ball, m₂ = 6.0 kg
initial speed of the first ball, u₁ = 12 m/s
initial speed of the second ball, u₂ = 4 m/s
let v be the final velocity of the two balls after the inelastic collision
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 12 + 6 x 4 = v(2 + 6)
48 = 8v
48 / 8 = v
v = 6 m/s
The initial kinetic energy of the balls is calculated as;
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²
K.E₁ = 144 + 48
K.E₁ = 192 J
The final kinetic of the balls is calculated as;
K.E₂ = ¹/₂(m₁ + m₂)(v²)
K.E₂ = ¹/₂(2 + 6)(6²)
K.E₂ = ¹/₂(8)(6²)
K.E₂ = 144 J
The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J
Therefore, the kinetic energy lost in the collision is 48 J
Answer:
D and A
Explanation:
Hint The basic difference between them is that a bar magnet is a permanent magnet whereas an electromagnet is a temporary magnet. An electromagnet is formed when an electric current is passed through wires wound around soft metalcore. An electromagnet loses its magnetism once the current flow is stopped.
Answer:
laser pointer does not make use of total internal reflection
In transistor,
Emitter current is equal to the sum of base current and collector current.
Thanks!
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
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