Answer:
a= 0.22 m/s²
Explanation:
Given that
M = 3.5 kg
θ = 30°
m = 1 kg
μ= 0.3
The force due to gravity
F₁= M g sinθ
F₁=3.5 x 10 x sin 30
F₁= 17.5 N
F₂ = m g
F₂ = 1 x 10 = 10 N
The maximum value of the friction force on the incline plane
Fr = μ M g cosθ
Fr = 0.3 x 2.5 x 10 cos30°
Fr= 6.49 N
Lets take acceleration of the system is a m/s²
F₁ - F₂ - Fr = (M+m) a
17.5 - 10 - 6.49 = (3.5+1)a
a= 0.22 m/s²
Answer:
I = 69.3 μA
Explanation:
Current through the straight wire, I = 3.45 A
Number of turns, N = 5 turns
Diameter of the coil, D = 1.25 cm
Resistance of the coil,
Distance of the wire from the center of the coil, d = 20 cm = 0.2 m
The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.
Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil
Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345
ΔB = -0.000001725
Induced current,
E = -N (Δ∅)/Δt
Δ∅ = A ΔB
Area, A = πr²
diameter, d = 0.0125 m
Radius, r = 0.00625 m
A = π* 0.00625²
A = 0.0001227 m²
Δ∅ = -0.000001725 * 0.0001227
Δ∅ = -211.6575 * 10⁻¹²
E = -N (Δ∅)/Δt
Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms
I = E/R
I = 0.0000693 A
I = 69 .3 * 10⁻⁶A
I = 69.3 μA
Answer:
Option A. 57.14 Ω
Explanation:
From the question given above, the following data were obtained:
Resistor 1 (R₁) = 100 Ω
Resistor 2 (R₂) = 400 Ω
Resistor 3 (R₃) = 200 Ω
Equivalent Resistor (Rₚ) =?
The equivalent resistor in the above circuit can be obtained as follow:
1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃
1/Rₚ = 1/100 + 1/400 + 1/200
Find the least common multiple (lcm) of 100, 400 and 200. The result is 400. Divide 400 by 100, 200 and 400 respectively and multiply the result with the numerator as shown
1/Rₚ = (4 + 1 + 2)/400
1/Rₚ = 7/400
Invert
Rₚ = 400/7
Rₚ = 57.14 Ω