Using Newton’s second law, why do you think a cotton ball may not be used as a baseball in a baseball game.

The reaction force is F = -8 N. It is noted that the Earth's gravitational force pulls the toy down with a weight of 8.0 N. This problem utilizes Newton's third law of motion, which states that "for every action, there is an equal and opposite reaction." In a mathematical context, it is illustrated as follows: where the force exerted on one object by another is indicated. In this scenario, the force of Earth acting on the toy is the action force. Consequently, the reaction force is characterized as F = -8 N (the negative indicates that the reaction force is directed upward). In essence, the toy exerts an upward force of 8 N on the Earth, thus comprising the required solution.

F = π/4 ρ d² v²

Explanation:

The formula for force is mass multiplied by acceleration:

F = ma

Acceleration is defined as the change in velocity over the change in time:

F = m Δv / Δt

Since there is no rebound effect, Δv is equal to v.

F = m v / Δt

Mass can be calculated as density multiplied by volume:

F = ρ V v / Δt

Flow rate describes the volume per time:

F = ρ Q v

Flow rate is determined by velocity multiplied by the cross-sectional area:

F = ρ (v A) v

This simplifies to F = ρ A v²

The area of a circle is calculated as pi times the square of the radius, or as pi/4 times the diameter squared:

F = ρ (π/4 d²) v²

Hence, F = π/4 ρ d² v²

When the boat submerges completely in the pond, the water level of the pond rises.

True. Explanation: In this instance, the area of the graph represents the impulse. Impulse is defined as the change in an object's momentum. Moreover, it is also expressed as the product of the force acting on an object and the duration of the impact. When we graph the force against time, if the force remains constant, the resultant graph will take on a rectangular shape, and the area under that graph will equal the impulse's definition.

Answer:

Explanation:

The transitions occur as follows:

P₁ V₁ changes to 3P₁, V₁ (with constant volume) — first phase.

Subsequently, 3P₁,V₁ transitions to 3P₁, 5V₁ (with constant pressure) — second phase.

During the initial phase, the temperature must be escalated by a factor of 3. Therefore, if the starting temperature is T₁, then the ending temperature will be 3 T₁.

P₁V₁ = n R T₁, where n represents the number of moles of gas.

Thus, nRT₁ = P₁V₁.

The heat added at constant volume is given by n Cv (3T₁ - T₁),

= n x 5/3 R X 2T₁ (noting that for diatomic gas, Cv = 5/3 R).

= 10/3 x nRT₁

= 10/3 x P₁V₁.

In the second phase, the temperature must rise 5 times. Thus, if the initial temperature is 3T₁, then the final temperature will be 15 T₁.

The heat added at constant pressure in this scenario becomes:

= n Cp (15T₁ - 3T₁)

= n x 7/3 R X 12T₁ (for diatomic gases, Cp = 7/3 R).

= 28 x nRT₁

= 28 P₁V₁.