m = 4, p = 4
Note that when two variables are placed directly next to each other, you are multiplying
plug in 4 for both m and p
(4)(4) + 4
Multiply
4 x 4 = 16
Add
16 + 4 = 20
20 is your answer
hope this helps
Answer:
![\theta = 5.83\ rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%205.83%5C%20rad)
Step-by-step explanation:
given,
angular deceleration, α = -0.5 rad/s²
final angular velocity,ω_f = 0 rad/s
angular position, θ = 6.1 rad
angular position at 3.9 s = ?
now, Calculating the initial angular speed
![\omega_f^2 = \omega_i^2 + 2 \alpha \theta](https://tex.z-dn.net/?f=%5Comega_f%5E2%20%3D%20%5Comega_i%5E2%20%2B%202%20%5Calpha%20%5Ctheta)
![0 = \omega_i^2 - 2\times 0.5\times 6.1](https://tex.z-dn.net/?f=%200%20%3D%20%5Comega_i%5E2%20-%202%5Ctimes%200.5%5Ctimes%206.1)
![\omega_i = \sqrt{6.1}](https://tex.z-dn.net/?f=%5Comega_i%20%3D%20%5Csqrt%7B6.1%7D)
![\omega_i = 2.47\ rad/s](https://tex.z-dn.net/?f=%5Comega_i%20%3D%202.47%5C%20rad%2Fs)
now, angular position calculation at t=3.9 s
![\theta = \omega_i t + \dfrac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega_i%20t%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
![\theta =2.47\times 3.9 - \dfrac{1}{2}\times 0.5\times 3.9^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D2.47%5Ctimes%203.9%20-%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%200.5%5Ctimes%203.9%5E2)
![\theta = 5.83\ rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%205.83%5C%20rad)
Hence, the angular position of the wheel after 3.9 s is equal to 5.83 rad.
Answer:
P = 6 2/3 ft
Step-by-step explanation:
The formula for the perimeter of a rectangle of length L and width W is
P = 2(L + W).
Here the perimeter is P = 2(1 2/3 + 5/6) = 2(5/2 + 5/6)
= 2(15/6 + 5/6) = 2(20/6)
= 40/6 = 20/3
P = 6 2/3 ft