Because most people are different and also are putting on a show for social medi
Yes bc math, numbers and more
Answer:
t₂ = 3.89 s
Explanation:
given,
speed of car = 23 m/s
speed of motorcycle = 23 m/s
after time of 4 s distance between them is equal to = 53 m
motorcycle accelerates at = 7 m/s
time taken to catch up with car = ?
let t₂ be the time in which motorcycle catches car.
distance traveled by car in t₂ s
d = 23 t₂ + 53
distance traveled by motorcycle
using equation of motion
![s = ut + \dfrac{1}{2}at^2](https://tex.z-dn.net/?f=s%20%3D%20ut%20%2B%20%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
![s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2](https://tex.z-dn.net/?f=s%20%3D%2023%20t_2%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%207%20%5Ctimes%20t_2%5E2)
now, equating both the distances
![23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2](https://tex.z-dn.net/?f=23t_2%20%2B%2053%3D%2023%20t_2%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%207%20%5Ctimes%20t_2%5E2)
![t_2^2 = 15.143](https://tex.z-dn.net/?f=t_2%5E2%20%3D%2015.143)
t₂ = 3.89 s
time taken by the motorcycle to catch the car is equal to 3.89 s
Speed of the tip of the minute hand=V=0.0244 cm/s
Explanation:
The angular velocity of the minute hand is given by
![\omega= \frac{2\pi}{T}](https://tex.z-dn.net/?f=%20%5Comega%3D%20%5Cfrac%7B2%5Cpi%7D%7BT%7D%20%20)
T= time period of the minute hand=60 min=3600 s
so ω= 2 π/3600 rad/s
Now linear velocity v= r ω
r= radius of minute hand=14 cm
so v= 14 (2 π/3600)
V=0.0244 cm/s
Answer:
The answer is below
Explanation:
The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm
radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m
a) Initial angular velocity (
) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s
θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
angular acceleration (α) is:
![\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\](https://tex.z-dn.net/?f=%5Comega%5E2%3D%5Comega_o%5E2%2B2%5Calpha%20%5Ctheta%5C%5C%5C%5C19.67%5E2-50.13%5E2%3D2%5Calpha%28272.9%29%5C%5C%5C%5C19.67%5E2%3D50.13%5E2%2B2%5Calpha%28272.9%29%5C%5C%5C%5C2%5Calpha%28272.9%29%3D-2126.108%5C%5C%5C%5C%5Calpha%3D-3.89%5C%20rad%2Fs%5E2%5C%5C%5C%5C)
b)
c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
a) When it stops, the final angular velocity is 0. Hence:
![\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)} =112.4\ rev](https://tex.z-dn.net/?f=%5Comega%5E2%3D%5Comega_o%5E2%2B2%5Calpha%20%5Ctheta%5C%5C%5C%5C0%3D50.13%5E2%2B2%28-3.89%29%5Ctheta%5C%5C%5C%5C2%283.89%29%5Ctheta%3D50.13%5E2%5C%5C%5C%5C2%283.89%29%5Ctheta%3D2513%5C%5C%5C%5C%5Ctheta%3D323%5C%20rad%5C%5C%5C%5Crevolutions%3D%5Cfrac%7B%5Ctheta%7D%7B2%5Cpi%20r%7D%3D%5Cfrac%7B323%7D%7B2%5Cpi%280.4572%29%7D%20%20%3D112.4%5C%20rev)
θ = 323 rad