In a constant acceleration of 3m per second, after 10 seconds,
3 x 10 = 30
B. 30m/s is your answer
hope this helps :D
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
16/9 m/s^2
negative 4/3 m/s^2
14 m/s
the last one is too detailed to do in my head while on the bus; sorry
0.495 m/s
Explanation
the formula for the terminal velocity is given by:
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20%5Ctext%7Bwhere%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
then
Step 1
Let's find the mass

now, replace
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2%280.002kg%29%289.81%5Ctext%7B%20%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%7B%282%5Ccdot10%5E3%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E3%7D%29%280.0001m%5E2%290.8%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.03924%5Cfrac%7B%5Coperatorname%7Bkg%7Dm%7D%7Bs%5E2%7D%7D%7B0.16%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E%7B%7D%7D%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B0.2452%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%20%5C%5C%20v%3D0.495%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
hence, the answer is 0.495 m/s