On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before sho

Refer to the attached file for the solution

Answer:

wow, still in Canada, by the way, my answer is ayy

History of the Philippines

The Philippines is a nation located in Southeast Asia near the Pacific Ocean. It comprises over 7,000 islands, serving as home to Filipinos. Here are several ways in which the strategic location of our country has contributed to shaping our history:

History of the PhilippinesThe Philippines is a nation located in Southeast Asia near the Pacific Ocean. It comprises over 7,000 islands, serving as home to Filipinos. Here are several ways in which the strategic location of our country has contributed to shaping our history:Being surrounded by bodies of water made fishing the primary livelihood of early Filipinos.

Over 7,000 islands, serving as home to Filipinos. Here are several ways in which the strategic location of our country has contributed to shaping our history:Being surrounded by bodies of water made fishing the primary livelihood of early Filipinos.When Ferdinand Magellan sailed, he landed on one of the islands in the Philippines. This marked the beginning of colonizationDue to the strategic location...

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

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:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + a

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + (2.25 x )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

= + 2gs

= + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying = + 2gs

= + 2(9.81 x 328)

= 0 + 6435.36

v = = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec