Today, we know that the continents rest on massive slabs of rock called tectonic plates. The plates are always moving and interacting in a process called plate tectonics. The continents are still moving today. Some of the most dynamic sites of tectonic activity are seafloor spreading zones and giant rift valleys.
A black hole doesn't suck things in any harder than any other object does, unless you're closer to it. The Earth hasn't gotten any closer to the sun in many years. This is not the reason for climate change.
Here is the answer. The microscope parts that should routinely be adjusted to control the light source and provide optimal illumination of the specimen are the following: <span>light source; condenser; specimen; objective lens; ocular lens. Hope this answers your question.</span>
Answer:
Bi. Current in 15.4 Ω (R₁) is 7.14 A.
Bii. Current in 21.9 Ω (R₂) is 5.02 A.
Biii. Current in 11.7 Ω (R₃) is 9.40 A.
C. Total current in the circuit is 21.56 A.
Explanation:
Bi. Determination of the current in 15.4 Ω (R₁)
Voltage (V) = 110 V
Resistance (R₁) = 15.4 Ω
Current (I₁) =?
V = I₁R₁
110 = I₁ × 15.4
Divide both side by 15.4
I₁ = 110 / 15.4
I₁ = 7.14 A
Therefore, the current in 15.4 Ω (R₁) is 7.14 A.
Bii. Determination of the current in 21.9 Ω (R₂)
Voltage (V) = 110 V
Resistance (R₂) = 21.9 Ω
Current (I₂) =?
V = I₂R₂
110 = I₂ × 21.9
Divide both side by 21.9
I₂ = 110 / 21.9
I₂ = 5.02 A
Therefore, the current in 21.9 Ω (R₂) is 5.02 A
Biii. Determination of the current in 11.7 Ω (R₃)
Voltage (V) = 110 V
Resistance (R₃) = 11.7 Ω
Current (I₃) =?
V = I₃R₃
110 = I₃ × 11.7
Divide both side by 11.7
I₃ = 110 / 11.7
I₃ = 9.40 A
Therefore, the current in 11.7 Ω (R₃) is 9.40 A.
C. Determination of the total current.
Current 1 (I₁) = 7.14 A
Current 2 (I₂) = 5.02 A
Current 3 (I₃) = 9.40 A
Total current (Iₜ) =?
Iₜ = I₁ + I₂ + I₃
Iₜ = 7.14 + 5.02 + 9.40
Iₜ = 21.56 A
Therefore, the total current in the circuit is 21.56 A
Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = 
- d fi / dt
= B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A 
Dt = - A 
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
