Answer:
θ=19.877⁰
Explanation:
Given data
Velocity Va=34.0 km/h
Velocity Va=100 km/h
To find
Angle θ
Solution
We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got
Sinθ=(Va/Vb)
Sinθ=(34.0/100)
θ=Sin⁻¹(34.0/100)
θ=19.877⁰
Find the horizontal components vcos30 ...one goes right and one goes left so they cancel each other.
Find vertical components vsin30.....there are two of them.... so 2vcos30....hey presto... resultant velocity = 2vCos30
Answer:
![P = 133.13 Watt](https://tex.z-dn.net/?f=P%20%3D%20133.13%20Watt)
Explanation:
Initial angular speed of the ferris wheel is given as
![\omega_i = 2\pi f](https://tex.z-dn.net/?f=%5Comega_i%20%3D%202%5Cpi%20f)
![\omega_i = 2\pi(8.5/3600)](https://tex.z-dn.net/?f=%5Comega_i%20%3D%202%5Cpi%288.5%2F3600%29)
![\omega_i = 0.015 rad/s](https://tex.z-dn.net/?f=%5Comega_i%20%3D%200.015%20rad%2Fs)
final angular speed after friction is given as
![\omega_f = 2\pi f](https://tex.z-dn.net/?f=%5Comega_f%20%3D%202%5Cpi%20f)
![\omega_f = 2\pi(7.5/3600)](https://tex.z-dn.net/?f=%5Comega_f%20%3D%202%5Cpi%287.5%2F3600%29)
![\omega_f = 0.013 rad/s](https://tex.z-dn.net/?f=%5Comega_f%20%3D%200.013%20rad%2Fs)
now angular acceleration is given as
![\alpha = \frac{\omega_f - \omega_i}{\Delta t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%20-%20%5Comega_i%7D%7B%5CDelta%20t%7D)
![\alpha = \frac{0.015 - 0.013}{15}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B0.015%20-%200.013%7D%7B15%7D)
![\alpha = 1.27 \times 10^{-4} rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%201.27%20%5Ctimes%2010%5E%7B-4%7D%20rad%2Fs%5E2)
now torque due to friction on the wheel is given as
![\tau = I \alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I%20%5Calpha)
![\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%286.97%20%5Ctimes%2010%5E7%29%281.27%20%5Ctimes%2010%5E%7B-4%7D%29)
![\tau = 8875.3 N m](https://tex.z-dn.net/?f=%5Ctau%20%3D%208875.3%20N%20m)
Now the power required to rotate it with initial given speed is
![P = \tau \omega](https://tex.z-dn.net/?f=P%20%3D%20%5Ctau%20%5Comega)
![P = 8875.3 \times 0.015](https://tex.z-dn.net/?f=P%20%3D%208875.3%20%5Ctimes%200.015)
Answer:
6.1 x 10^-8 newtons
Explanation:
F = 8.98 *109 *1*1/3845000002
Same magnitude (1000N) and opposite direction
you may also answer -1000 N