The equation can be formed as (x - 2)(x + 3) = 0, which expands to x^2 + x - 6 = 0.

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). 1 * sin(30°) = 2 * sin(θ2) => sin(θ2) = 0.25 => θ2 = 15°.

A ray of light passing through the optical center of a lens continues in a straight line without bending.

A ΔG of +5 kJ/mol indicates that the reaction is non-spontaneous.

ΔG = ΔH - TΔS = 100 kJ - 298 K * (-0.2 kJ/K) = 100 kJ + 59.6 kJ = 159.6 kJ.

Using Ohm's Law, V = I * R = 2 A * 5 Ω = 10 V.

Using Ohm's Law, R = V / I = 9 V / 3 A = 3 Ω.

The rotational kinetic energy is given by (1/2)Iω^2, where I is the moment of inertia.

The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which can be expressed as (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2).

The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which simplifies to (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.

The linear velocity v of a rolling object is given by v = Rω.

For rolling without slipping, the relationship is v = ωR, where v is the translational speed and ω is the angular speed.

Angular momentum L = Iω, thus ω = L/I.

Angular momentum L = Iω = 4 kg·m² * 3 rad/s = 12 kg·m²/s.

Angular momentum L is given by L = Iω. Thus, L = 5 kg·m² * 3 rad/s = 15 kg·m²/s.

Angular momentum L = Iω = 5 kg·m² * 3 rad/s = 15 kg·m²/s.

The rotational kinetic energy is given by KE = 1/2 Iω^2.

A geostationary satellite has an orbital period equal to the Earth's rotation period, which is 24 hours.

In a stable orbit, the net force acting on the satellite is zero because the gravitational force provides the necessary centripetal force.

The potential energy of a satellite increases when it is moved to a higher orbit due to the increase in distance from the Earth's center.

The orbital period of a satellite increases when it is moved to a higher orbit, according to Kepler's third law.

For a satellite in a stable circular orbit, the centripetal acceleration is equal to the gravitational acceleration.

A satellite in a circular orbit possesses both kinetic energy due to its motion and potential energy due to its position in the gravitational field.

If a satellite's altitude is doubled, its orbital speed decreases by √2.

As the altitude increases, the orbital period increases due to the greater distance from the Earth's center.

If a satellite's speed exceeds the escape velocity, it will escape Earth's gravitational pull and move away into space.

If a satellite's speed is less than the required orbital speed, it will not have enough centripetal force to maintain its orbit and will fall back to Earth.

a_5 = 2(5) + 3 = 13.

Time period (T) is the reciprocal of frequency (f). T = 1/f = 1/1 = 1 s.

The amplitude of a simple harmonic oscillator is defined as the maximum displacement from the equilibrium position, which is 5 cm in this case.