Answer:
The amount of work is 89600 ft-lb.
Explanation:
Given that,
Length of the cable = 160 ft
Weight density = 7 lb/ft
Weight of cable to be filled = 7(160-x)
We need to calculate the amount of work
Using formula of work done
![\int{dW}=\int_{0}^{160}{7(160-x)}dx](https://tex.z-dn.net/?f=%5Cint%7BdW%7D%3D%5Cint_%7B0%7D%5E%7B160%7D%7B7%28160-x%29%7Ddx)
On integration
![W=[1120x-\dfrac{7x^2}{2}]_{0}^{160}](https://tex.z-dn.net/?f=W%3D%5B1120x-%5Cdfrac%7B7x%5E2%7D%7B2%7D%5D_%7B0%7D%5E%7B160%7D)
![W=1120\times160-0-\dfrac{7\times(160)^2}{2}+0](https://tex.z-dn.net/?f=W%3D1120%5Ctimes160-0-%5Cdfrac%7B7%5Ctimes%28160%29%5E2%7D%7B2%7D%2B0)
![W=89600\ ft-lb](https://tex.z-dn.net/?f=W%3D89600%5C%20ft-lb)
Hence, The amount of work is 89600 ft-lb.
<u>Answer:</u>
Horizontal component of the velocity when the opposing player fields the ball = 11.00 m/s
<u>Explanation:</u>
The velocity of a body in 2 dimension can be resolved in to 2 parts, horizontal and vertical component. In the case of free falling or projectile body the horizontal component remains the same but vertical component is affected by acceleration due to gravity.
In this case Initial velocity = 17.6 m/s
Angle between horizontal axis and direction of velocity = 51.3 °
We know that horizontal component = v cos θ
Vertical component = v sin θ
Since the horizontal component remains the same, it is unchanged when the opposing player fields the ball.
So horizontal component of the velocity = v cos θ = 17.6 * cos 51.3 °
= 11.00 m/s