B. Exfoliation. Hope I helped you out bro.
The calculated mutual inductance is 8.544 x 10⁻⁵ H.
Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.
Length of the solenoid= 5.0 cm
Area of cross-section=1.0 cm²
no of spaced turns=300 turns
turns of insulated wire=180 turns
Mutual inductance (M) = μ₀μr N1N2 A/ L
=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5
=79.12 x 10⁻¹⁰ x 54000 / 5
=8.544 x 10⁻⁵ H
hence, the mutual inductance is 8.544 x 10⁻⁵ H.
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Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²
Friction , as the angle gets huger and higher , this is less and less normal force into the inclined plane .