3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the

<span>a. To determine the velocity at which the camera strikes the ground: v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera impacts the ground with a speed of 42 m/s. b. To calculate the duration it takes for the camera to reach the bottom: y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds
       
The camera descends for 11.4 seconds before hitting the ground.</span>

Answer:

The duration, t = 3.53 seconds

Explanation:

The following information is provided:

The equation to calculate the time t is expressed as:

...... (1)

Where

s denotes the distance in feet

We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet

Substituting the value of s into equation (1) yields:

t = 3.53 seconds

Thus, the time taken by the object is 3.53 seconds, which provides the required answer.

Response:

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

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