A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

Conversely, in darkness, the diameter typically enlarges to 7 mm.

This indicates an increase in light-gathering capacity.

Thus, the amount of light the eye can capture is 31.4 mm².

Answer:

(A) = 3.57 m

Explanation:

According to the question, the information provided is:

diameter (d) = 3.2 m

mass (m) = 42 kg

angular speed (ω) = 4.27 rad/s

Using the conservation of energy principle, we have

mgh = 0.5 mv² + 0.5Iω²...equation 1

where

Inertia (I) = 0.5mr²

ω = v/r

Revising equation 1, it turns into

mgh = 0.5 mv² + 0.5(0.5mr²)(v/r)²

resulting in gh = 0.5 v² + 0.5(0.5)v²

This simplifies to 4gh = 2v² + v²

thus h = 3v² ÷ 4g... equation 2

Given ω = v/r, we find v = ωr = 4.27 × (3.2 ÷ 2)

which yields v = 6.8 m/s

Next, substituting the value of v into equation 2 gives us

h = 3v² ÷ 4g

h = 3 × (6.8)² ÷ (4 × 9.8)

h = 3.57 m

Response:

The normal force acting on the ant is 0.75 N.

Explanation:

Provided;

the diameter of the ball, D = 40 cm = 0.4 m

radius of the ball, r = 0.2 m

weight of the beach ball, m₁ = 300 g = 0.3 kg

weight of the ant, m₂ = 4 x 10⁻⁶ kg

velocity of the ball, v = 4 m/s

The area of a spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) encountered by the spherical ball is given as;

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is the air density = 1.21 kg/m³

The downward force from the ball caused by its weight and that of the ant is given by;

The resultant downward force acting on the ball is given by;

This downward force is equal to the normal reaction it applies to the ant.

Consequently, the normal force impacting the ant is 0.75 N.

Answer:

a) n = 1.33, b)  t = 87 10⁻⁹ m

Explanation:

In Part A.

The very thin anti-reflection layer needs to induce destructive interference for the targeted wavelength.

      2t sin θ = (m + ½) λₙ

The factor of 2 comes from the path of light inside the film, where λₙ is the wavelength adjusted for the film's refractive index

      λₙ =  λ₀ / n

      2t = (m + ½) λ₀ / n

      n = (m+ ½) λ₀/2t

Assuming we are evaluating the first interference order, where m = 0

     n = (½ 480 10-9)/ 2 90

     n = 1.33

In Part B

We may not have the exact refractive index for the glass, but it generally approximates around 1.5; since the film's index is less, it incurs no phase alteration.

      2t = (m + ½) λ₀ / n

      t = 1/4 480 10⁻⁹ / 1.38

      t = 87 10⁻⁹ m