A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b

I’ve provided the missing image. We can analyze this scenario by applying the principle of energy conservation. At point A, the car possesses both potential and kinetic energy. As it moves down the track, some initial energy is lost due to friction. Thus, as it approaches point B, we have a specific amount of energy remaining. According to the conservation of energy laws, this remaining energy at point B will equal the sum of its kinetic and potential energy.

Answer:

Height (h) = 17 m

Velocity (v) = 18.6 m/s

Explanation: This problem can be solved using kinematic motion equations.

Given Data

Initial velocity (u) = 0

Acceleration (a) = g

Time (t) = 1.9 seconds

First, we calculate the height.

Then, we find the final velocity

The acceleration graph is a linear representation described by y=9.8, as it remains constant:

The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):

The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):

These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.

Answer:

Explanation:

Provided:

mass of the steel ball

initial velocity of the ball

Final velocity of the ball (moving upwards)

The impulse given is determined by the change in the momentum of the object.

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Thus, the magnitude of the Impulse is 4 N-s.

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Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

I = Ce^kt

At t=0, I=I(0) implies C=I(0)

I = I(0)e^kt

At t=3 & I=0.25I(0), we find 0.25=e^3k

Solving for k gives k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

The intensity of light 18 feet underwater is about 0.2%