Question

A turntable rotates with a constant angular acceleration 2.30rad/s^2 . After a time interval 3.75s the wheel has rotated through an angle of 59.0rad .


What was the angular velocity of the wheel at the beginning of the time interval 3.75s ?

Answer 1

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Answer 2

Do..u have answer key with u ....@michelle092294

Answer 3

no i dont:( this is online assignment

Answer 4

oh i got it:)

Answer 5

wait its still wrong -_-

Answer 6

ill try

Answer 7

wrong

Answer 8

lol

Answer 9

\[\alpha = \frac{ d \omega }{ dt }\]

Answer 10

angular accl = change in angular velocity

Answer 11

\[2.3 * 3.75 = \delta \omega\]

Answer 12

\[\omega f - \omega i = 8. 625\]

Answer 13

wrong

Answer 14

lol....not finished Yet....

Answer 15

\[\omega i = \omega f - 8.625\]

Answer 16

\[\omega f = 56 / 3.75 = 15.73\]

Answer 17

wrong.

Answer 18

yeah. its wrong . i try that answer awhile ago -_-

Answer 19

\[\omega i = 15.73 - 8.625 = 7.10\]

Answer 20

lol.....Guys wait..) let me Finish

Answer 21

more wrong.

Answer 22

sure please finish

Answer 23

Do u have options ..)

Answer 24

nope. i dont have

Answer 25

Hm....... My appro Answer is 7.10

Answer 26

ill try

Answer 27

wrong -_-

Answer 28

hm.....)

Answer 29

help me :(

Answer 30

\[\large 59=\omega_i \cdot 3.75+\frac12 \cdot 2.30\cdot (3.75)^2\]

\[\Large \omega_i= {59 \over 3.75} - \frac 12 \cdot 2.30 \cdot3.75\]

Answer 31

thanks :)

Answer 32

\[\approx 11.4 radians /second\]

Answer 33

got it :)