The second law states that the total entropy can never decrese over time for an isolated system
Answer:
Velocity of both masses after the collisio
Explanation:
Hope it will help
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<em><u>Brainlists please</u></em></h2>
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Answer:
1.08x10⁻⁷
Explanation:
F=(GM₁M₂)/r²
=((6.67x10⁻¹¹)(70)(52))/(1.5²)
=2.42788x10⁻⁷/2.25
=1.07905778x10⁻⁷
≈1.08x10⁻⁷
Answer:
7.78 * 10³ m/s
Explanation:
Orbital velocity is given as:
v = √(GM/R)
G = 6.67 * 10^(-11) Nm/kg²
M = 5.98 * 10^(24) kg
R = radius of earth + distance of the satellite from the surface of the earth
R = 2.15 * 10^(5) + 6.38 * 10^(6)
R = 6.595 * 10^(6) m
v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))
v = √(6.048 * 10^7)
v = 7.78 * 10³ m/s
