By definition of circumference, the length of the arc EF (radius: 6 in, central angle: 308°) shown in red is approximately equal to 32.254 inches.
<h3>How to calculate the length of an arc</h3>
The figure presents a circle, the arc of a circle (s), in inches, is equal to the product of the <em>central</em> angle (θ), in radians, and the radius (r), in inches. Please notice that a complete circle has a central angle of 360°.
If we know that θ = 52π/180 and r = 6 inches, then the length of the arc CD is:
s = [(360π/180) - (52π/180)] · (6 in)
s ≈ 32.254 in
By definition of circumference, the length of the arc EF (radius: 6 in, central angle: 308°) shown in red is approximately equal to 32.254 inches.
<h3>Remark</h3>
The statement has typing mistakes, correct form is shown below:
<em>Find the length of the arc EF shown in red below. Show all the work.</em>
To learn more on arcs: brainly.com/question/16765779
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All i got was 1 and a half months
we know that
The scalar magnitude of the velocity vector is the speed. The speed is equal to

in this problem we have

substitute in the formula


therefore
<u>the answer Part a) is</u>
the speed is equal to 
<u>Part b) </u>Find the velocity
we know that
<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it
in this problem we have
the magnitude is equal to the speed


therefore
<u>the answer Part b) is</u>
the velocity is 
Part c)
we know that
the acceleration is equal to the formula

in this problem we have


substitute in the formula



therefore
<u>the answer Part c) is</u>
the acceleration is 
This is an example of negative acceleration
the volume of a sphere can be calculated with the formula v=4/3(pi)(r)^3