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.
Hi there! So 39,300 copies of a book were sold on debut month of release, and that represents 6.3% of all copies sold to date. To find the total amount of copies sold, we can write and solve a proportion. Set it up like this:
39,300/x = 6.3/100
We set it up like this because 39,300 is part of the total amount, and it represents 6.3% of the total book sales. Percents are parts of 100, which is why 6.3 is above 100. Let's cross multiply the values. 39,300 * 100 is 3,930,000. 6.3 * x is 6.3x. that makes 3,930,000 = 6.3x. Divide each side by 6.3 to isolate the x. 6.3x/6.3 cancels out. 3,930,000/6.3 is 623,809.5238 or 623,810 when rounded to the nearest whole number. There. The total amount of copies sold to date is about 623,810.
Answer:
a
Step-by-step explanation:
its a
Let's draw!
H 1 T
HT 2 HT
HTHT 3 HTHT
HTHTHT 4 HTHTHT
HTHTHTHT 5 HTHTHTHT
HTHTHTHTHT 6 HTHTHTHTHT
You can count the probabilities using this.
HHHHTT
HHHTHT
HHTHHT
HHTHTH
HHHTTH
HTHHHT
HTTHHH
HTHTHH
HTHHTH
THHHHT
THHHTH
THHTHH
THTHHH
TTHHHH
Therefore, I think the probability is 14/64. Not sure so check my work.
14 combinations
64 outcomes (2+4+8+16+32+64 or 2^n)
Answer:
1/ 5 and 2/3
cross multiple
example
1 2 = 10
/ x / = /
5 3 3
= 10/3
so it's 3 1/3 + 10 Mm
so may guess is
13.3333 infinite or 13.34 M
Dont know if its right but good luck