The energy needed is 35.4 KJ, to pop the corn of a given quantity, and by using latent heat of vaporization of water.
What is Energy?
When work is done and heat and light are produced, energy is the quantitative quality that is transferred to the body or to the physical system. The law of conservation of energy asserts that although energy can be changed in form, it cannot be created or destroyed. Energy is a conserved quantity.
Calculations:
The boiling point of water is 175 °C.
The mass of corn is 95.0 g.
The percentage of water in corn is 14%.
The latent heat of vaporization for water at 175 °C is 0.90 times its value at 100 °C.
Calculate the mass of water,
Mw = (14/100) × (95g) × (1kg/1000g)
= 0.0133 Kg
calculate the latent heat of vapourization,
Lv = (0.9) × (2.26 × [tex]10^{6}[/tex]J/kg)
= 2.034 × [tex]10^{6}[/tex] J/kg
Total heat required,
Q = Mw Lv + Mw. C Δ T
Substitute the values,
Q = (0.0133) × (2.034 × [tex]10^{6}[/tex]) + (0.0133) × (4186) × (175.25)
= (35403.27 J) × ( 1KJ/ 1000J)
= 35.4 KJ
Hence, the energy needed is 35.4 KJ, to pop the corn of a given quantity.
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A white ball traveling at 5.0 m/s east hits a stationary orange ball that has
the same mass. The white ball stops, and the orange ball moves away at 5.0
m/s east. What is true for this collision?
The collision is inelastic because kinetic energy is conserved.
A
B
The collision is inelastic because kinetic energy is not conserved.
C The collision is elastic because kinetic energy is conserved.
D The collision is elastic because kinetic energy is not conserved.
The kinetic energy of balls is conserved, the collision is elastic.
What is conservation of energy?Energy conservation in physics is a fundamental law of chemistry and physics stating that the total energy in an isolated system is constant despite internal changes. It is most commonly expressed as "energy can neither be created nor destroyed", and is the basis of the first law of thermodynamics.An elastic collision is one in which the system suffers no net loss of kinetic energy as a result of the collision. In an elastic collisions, both momentum and kinetic energy are conserved.When one reactant loses momentum or kinetic energy, the other reactant gains the very same amount of momentum or kinetic energy. Thereby, the total amount of kinetic momentum and energy of the system would then remain as it is, and therefore, kinetic energy is conserved.To learn more about kinetic energy refer to :
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when an object is placed 12 inches from the face, how will the lens change to help focus on the object?
When an object is placed 12 inches from the face, then the lens : will thicken and therefore refract light more strongly.
How does the lens change to focus on objects that are close?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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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.
Free fall motion is motion in which the only force acting on the body is gravity.
What is free fall motion?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.