Horizontal component (vx) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Time of flight (T) = (2u * sin(θ)) / g = (2 * 40 * (√3/2)) / 9.8 ≈ 6.1 s.

Horizontal component (vx) = u * cos(θ) = 40 * 0.5 = 20 m/s.

Horizontal component (vx) = u * cos(θ) = 50 * cos(30) = 50 * (√3/2) = 43.3 m/s.

Horizontal component (v_x) = u * cos(θ) = 50 * 0.5 = 25 m/s.

Horizontal component (u_x) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Time of flight (T) = (2u * sin(θ)) / g = (2*40*sin(45))/9.8 = 5.14 s.

Vertical component (u_y) = u * sin(θ) = 20 * (√2/2) = 14.14 m/s.

The magnetic force is proportional to the velocity of the charge, so if the velocity is doubled, the magnetic force also doubles.

Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), i = 30 degrees, n2 = 1.5 (glass). Thus, sin(r) = (1 * sin(30))/1.5 = 0.333, giving r ≈ 18.4 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 45 degrees. Solving gives r = 30 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1, n2 = 1.5, i = 30 degrees. Solving gives r ≈ 18.4 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so 1 * 0.5 = 1.33 * sin(θ2). Solving gives θ2 ≈ 22.5 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water), θ1 = 30 degrees. Thus, sin(θ2) = (1 * sin(30))/1.33 = 0.375. Therefore, θ2 = sin^(-1)(0.375) which is approximately 22 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2), we find θ2 = sin^(-1)(sin(30 degrees)/1.33) = 22.5 degrees.

Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water). Solving gives θ2 ≈ 22 degrees.

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so sin(θ2) = (1 * 0.5) / 1.33 = 0.375. Thus, θ2 = sin^(-1)(0.375) ≈ 22.5 degrees.

When a ray of light passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.

When a ray passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.

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

According to the law of reflection, the angle of reflection is equal to the angle of incidence, which is 30 degrees.

According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is 30 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). For air (n1=1) and water (n2=1.33), we find r = 45 degrees.

The moment of inertia of a rectangular plate about an edge is I = 1/12 Ma^2 + 1/3 Mb^2.

Using Ohm's Law, I = V/R = 20V / 10Ω = 2 A.