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
Mass of the block = 1.4 kg
Weight of the block = mg = 1.4 × 9.8 = 13.72 N
Normal force from the surface (N) = 13.72 N
Acceleration = 1.25 m/s^2
Let the coefficient of kinetic friction be μ
Friction force = μN
F(net) = ma
μmg = ma
μg = a
μ = 
μ = 
μ = 0.1275
Hence, the coefficient of kinetic friction is: μ = 0.1275
29.5 days
It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).
Answer:
Making a quick cut left to intercept a pass
Explanation:
It takes more energe to do than running
