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Answer:
5.76 μC
Explanation:
The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T
So, ε = -ΔΦ/Δt
ε = -NAΔB/Δt
ε = -NAΔB/Δt
Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)
So, iR = -NAΔB/Δt
iΔt = -NAΔB/R
Δq = -NAΔB/R where Δq = charge = iΔt
substituting the values of the variables into the equation, we have
Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω
Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω
Δq = 0.011π × 10⁻¹ m²T/600 Ω
Δq = 0.03456 × 10⁻¹ m²T/600 Ω
Δq = 5.76 × 10⁻⁶ C
Δq = 5.76 μC
Answer:
"The properties of magnets are used to make electricity. ... Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy."
Source: https://www.eia.gov/energyexplained/electricity/magnets-and-electricity.php
Answer:
The scientific notation for the each number is given below:
Explanation:
The scientific notation for the following numbers are:
a. For 0.00000123 N
1.23 × 10^-6
b. For 417 000 000 kg
4.17 × 10^8
c. For 246800 A
2.47 × 10^5
d. For 0.00088m
8.8 × 10^-4
The above represent scientific notations of each numbers
consider the motion of the mass parallel to the incline
v₀ = initial velocity at the bottom of incline = 0 m/s
v = final velocity at the top of incline = 8.00 m/s
a = acceleration
d = displacement = L = length of incline = 15 m
using the equation
v² = v²₀ + 2 a d
8² = 0² + 2 a (15)
64 = 30 a
a = 64/30
a = 2.13 m/s²
F = applied force
from the force diagram, perpendicular to incline , force equation is given as
N = mg Cos30
μ = Coefficient of friction = 0.426
frictional force acting on the mass is given as
f = μ N
f = μ mg Cos30
parallel to incline , force equation is given as
F - f - mg Sin30 = ma
F - μ mg Cos30 - mg Sin30 = ma
inserting the values
F - (0.426 x 40 x 9.8) Cos30 - (40 x 9.8) Sin30 = 40 (2.13)
F = 425.82 N