An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

where:

represents the pressure difference across the two ends

denotes the viscosity of the fluid

L signifies the length of the pipe

indicates the volumetric flow rate, where is the cross-sectional area of the tube and refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

is the dynamic viscosity of water at

L=4.0 cm=0.04 m

and r=1 mm=0.001 m

Substituting these values into the formula yields:

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.

Answer:

C) True. The distance S increases over time, with v₁ = gt and v₂ = g (t-t₀), illustrating that v₁> v₂ for the same t.

Explanation:

We have a set of statements to evaluate for correctness. The most effective approach is to examine the problem in detail.

Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Stone 2

Released shortly thereafter, let's assume a delay of one second, we can utilize the same timing mechanism

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

We can now calculate the separation distance between the two stones, which is applicable for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t² - 2 t t₀ + t₀²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t - t₀)

This represents the distance between the two stones over time, with the coefficient outside the parentheses being constant.

For t < to, the first stone remains stationary while the distance grows.

For t > = to, the expression (2t/to-1) yields a value greater than 1, indicating that the distance expands as time progresses.

We can now analyze the different statements

A) false. The height difference increases over time.

B) False S increases.

C) It is true that S increases over time, with v₁ = gt and V₂ = g (t-t₀) indicating v₁> v₂ at the same t.

Answer:

The car that is the furthest from the finish line is: Car III (Choice C).

Explanation:

Here, we seek the car with the lowest overall average speed throughout the race. Thus, the one in last place inherently possesses the slowest average speed.

Since Car III is significantly behind Cars I and II, Choice A and B cannot be correct. Choice D is also not valid, as the positions of the cars are not the same. Lastly, Choice E is incorrect due to sufficient evidence demonstrating that Choice C has the lowest average speed.