Answer:
4E
Explanation:
From the question given above, the following data were obtained:
Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m
Initial energy (E₁) = E
Final elongation (e₂) = 0.04 + 0.04 = 0.08 m
Final energy (E₂) =?
The energy stored in a s spring is given by:
E = ½Ke²
Where
E => is the energy
K => is the spring constant
e => is the elongation
From:
E = ½Ke²
Energy is directly proportional to the elongation. Thus,
E₁/e₁² = E₂/e₂²
With the above formula, we can obtain the final energy as follow:
Initial elongation (e₁) = 0.04 m
Initial energy (E₁) = E
Final elongation (e₂) = 0.08 m
Final energy (E₂) =?
E₁/e₁² = E₂/e₂²
E / 0.04² = E₂ / 0.08²
E / 0.0016 = E₂ / 0.0064
Cross multiply
0.0016 × E₂ = 0.0064E
Divide both side by 0.0016
E₂ = 0.0064E / 0.0016
E₂ = 4E
Therefore, the final energy is 4 times the initial energy i.e 4E
Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:
Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:
Using the arcsin properties, we get:
Obs: Approximate results, and the drawing is attached
If you notice any mistake in my english, let me know, because i am not native.
Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
Answer:
high frequent sound
Explanation:
because if its low than its slower.
