hw 27 question 2 part d The text reads:"A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 33 s."There are no graphs or diagrams in the image. **Problem Statement:**Find the magnitude of a passenger's acceleration at the bottom of the wheel.**Instructions:**Express your answer using two significant figures.**Answer Input Box:**- The input box is labeled with the variable \( a \) and the unit m/s\(^2\).**Buttons and Options:**- There are several buttons located above the input box: - A button for entering an equation or using special symbols. - A button marked with a question mark for help or hints. - Undo and redo arrows for correcting entries. - A keyboard icon suggesting additional input options.**Submission Options:**- A "Submit" button to enter your answer.- A "Request Answer" link for additional assistance or solutions.

Name: hw 27 question 2 part d The text reads:"A Ferris wheel with a radius o
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: hw 27 question 2 part d The text reads:"A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 33 s."There are no graphs or diagrams in the image. **Problem Statement:**Find the magnitude of a passenger's ac

A Person sitting on t a Ferry wheel experiences the force of gravity and the Centrepetal force both acting downwards in the same direction. The circumference of a cirlce is given by the equation, C=2πr ......(1) Where r is the radius of the cirlce. Velocity of a body can be determined using the relation, v=dt ......(2) Where, d and t are the distance and time respectively. The Centripetal force 'Fc'is given by the formula, Fc=mv2r ......(3) Where m is the mass of the body, v is the velocity, and r is the radius of the circular motion. The circumference of the Ferry wheel will be given by equation(1) as, C=2πr Substitute the radius of the ferry wheel, r=9.5 m ⇒C=23.14(9.5 m) =59.66 m The Ferry wheel completes one revolution every 33 seconds. Its velocity can be determined by the formula, v=dt Substitute, the distance d=59.66 m and time t = 33 s. v=59.66 m33 s =1.81 m/s The Centripetal force experienced by the person sitting on the ferris wheel will be given by equation (3) as, Fc=mv2r The Free body diagram for the forces acting on the person sitting at the bottom of the Ferry wheel is, From the Free body diagram, the force acting downward is the sum of Centripetal force and the gravitational force. ∑F=mg+mv2r⇒ma=mg+mv2r ⇒a=g+v2r Substitute the known values, Acceleration due to gravity g=9.8 m/s2, Velocity calculated from the previous step v=1.81 m/s, and the radius of the ferry wheel, r = 9.5 m. a=9.8 m/s2+(1.81 m/s)29.5 m =10.15 m/s2 Therefore the acceleration of the person on the bottom of the ferry wheel is 10.15 m/s2.