because of the pressure inside a popcorn kernel, water does not vaporize at 100°C. Instead, it stays liquid until its temperature is about 175°C, at which point the kernel ruptures and the superheated water turns into steam. How much energy is needed to pop 95.0 g of corn if 14 percent of a kernel's mass consists of water? Assume that the latent heat of vapor- ization for water at 175°C is 0.90 times its value at 100°C and that the kernels have an initial temperature of 175°C.​

When an object is placed 12 inches from the face, then the lens : will thicken and therefore refract light more strongly.

The process of changing the shape of the lens to focus on near or distant objects is called accommodation. To focus on a near object , lens becomes thicker which allows the light rays to refract more strongly.

Cornea provides most of the eye's optical power or light-bending ability. When the light passes through the cornea, it is bent again for more finely adjusted focus by the crystalline lens inside the eye. Lens helps in focusing the light on the retina.

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Free fall motion is motion in which the only force acting on the body is gravity.

A free-falling object is one that moves only due to the effect of gravity, and its motion is defined by Newton's second law of motion. We can use algebra to calculate the acceleration of a free-falling particle.

The student should launch the sphere at 2v₀, for the sphere will land at approximately 1.41D, which is in the 3 point zone

The given parameter are;

The distance covered by the sphere when launched at height, H = D

The velocity with which the ball reaches D = v₀

The current available height of launcher= H/2

The available velocities = v₀, and 2v₀

Now,

From H = ut + (1/2)gt², where, initial velocity of the vertical motion of the ball, u = 0

we know;

H = (1/2)gt²

Therefore, the time it takes the ball to drop from H, t = √(2H/g)

The distance, D = v₀ × √(2H/g)

When the height is H/2, we get;

t = √(2H/(2g)) = √(H/g)

Thus, the distance covered, D₁ = v₀ × √(H/g)

Therefore, D = (√2) × v₀ × √(H/g) = (√(2))D₁

Now, D₁ = D/(√2) ≈ 0.71D

D₁ ≈ 0.71D

At speed 2v₀, we have;

D₂ = 2v₀ × √(H/g) = √2 × v2 × v₀ × √(H/g) = √2 × v₀ × √(2H/g) = √2D₁ ≈ 1.41D

D₂ ≈ 1.41D

The 2 point zone = D/2 < x < D = 0.5D < x < D (Position D₁ ≈ 0.71D is located here)

The 3 Point Zone =  D < x < 3D/2 = D < x < 1.5D (Position D₂ ≈ 1.41D is located here)

Given that at D₂, the ball lands in the 3 Point Zone, the student should launch the sphere at the speed, 2v₀, so that the ball will land at D₂ ≈ 1.41D,  which is in the 3 Point Zone

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The complete question is as follows:

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizontal direction with variable speed and from a variable vertical position and a fixed horizontal position x=0.

The robot is calibrated by adjusting the speed at which the sphere is launched and the height of the robot’s sphere launcher. Depending on where the spheres land on the ground, students earn points based on the accuracy of the robot. The robot is calibrated so that when the spheres are launched from a vertical position y=H and speed v0, they consistently land on the ground on a target that is at a position x=D. Positive directions for vector quantities are indicated in the figure.

When the students arrive at the competition, it is determined that the height of the sphere launcher can no longer be adjusted due to a mechanical malfunction. Therefore, the spheres must be launched at a vertical position of y=H2. However, the spheres may be launched at speed v0 or 2v0.

Question: In a clear response that may also contain diagrams and/or equations, describe which speed, v0 or 2v0, the students should launch the sphere at so that they earn the maximum number of points in the competition.

The kinetic energy of balls is conserved, the collision is elastic.

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