Answer:
43 m\s in the 10 seconds?
Explanation:
Answer:
Imp = -4 [kg*m/s]
Explanation:
To determine the impulse we must know the initial speed and final velocity of the ball, that is before and after hitting the wall. As these data are given in the input information, we can very easily find the value of the impulse given during the hit on the wall.
![m*v_{1}+Imp=mv_{2} \\Imp= m*v_{2}-m*v_{1}\\Imp = (2*4)-(2*6)\\Imp = -4[kg*\frac{m}{s} ] \\](https://tex.z-dn.net/?f=m%2Av_%7B1%7D%2BImp%3Dmv_%7B2%7D%20%20%5C%5CImp%3D%20m%2Av_%7B2%7D-m%2Av_%7B1%7D%5C%5CImp%20%3D%20%282%2A4%29-%282%2A6%29%5C%5CImp%20%3D%20-4%5Bkg%2A%5Cfrac%7Bm%7D%7Bs%7D%20%5D%20%5C%5C)
The negative sign means that the impulse is given in the opposite direction to the one that the ball was thrown.
The angular acceleration is equal to the final angular velocity divided by the time and the average angular velocity is equal to half the final angular velocity. It follows that the rotational kinetic energy given to the flywheel is equal to the work done by the torque.
The wave property is called frequency
Answer
given,
mass of the gate = 4.5 kg
distance between the hinge 1.5 m
mass of the bird = 1.1 kg
velocity of the bird = 5 m/s
![I_{gate} =\dfrac{1}{3}ML^2](https://tex.z-dn.net/?f=I_%7Bgate%7D%20%3D%5Cdfrac%7B1%7D%7B3%7DML%5E2)
![I_{gate} =\dfrac{1}{3}\times 4.5 \times 1.5^2](https://tex.z-dn.net/?f=I_%7Bgate%7D%20%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%204.5%20%5Ctimes%201.5%5E2)
![I_{gate} =3.375 kg.m^2](https://tex.z-dn.net/?f=I_%7Bgate%7D%20%3D3.375%20kg.m%5E2)
a) from conservation of angular momentum
![m(v_1+v_2)\dfrac{W}{2}= 3.375 \omega_f](https://tex.z-dn.net/?f=%20m%28v_1%2Bv_2%29%5Cdfrac%7BW%7D%7B2%7D%3D%203.375%20%5Comega_f)
![1.1(5+2)\dfrac{1.5}{2}= 3.375 \omega_f](https://tex.z-dn.net/?f=%201.1%285%2B2%29%5Cdfrac%7B1.5%7D%7B2%7D%3D%203.375%20%5Comega_f)
![\omega_f= 1.711\ rad/s](https://tex.z-dn.net/?f=%5Comega_f%3D%201.711%5C%20rad%2Fs)
b) There is no external torque hence, momentum is conserved
initial kinetic energy = ![\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
initial kinetic energy = ![\dfrac{1}{2}\times 1.1 \times 5^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%201.1%20%5Ctimes%205%5E2)
= 13.75 J
final kinetic energy
KE = ![\dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cdfrac%7B1%7D%7B2%7DI%5Comega%5E2)
KE = ![\dfrac{1}{2}\times 1.1 \times 2^2 + \dfrac{1}{2}\times 3.375\times 1.711^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%201.1%20%5Ctimes%202%5E2%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%203.375%5Ctimes%201.711%5E2)
KE = 7.14 J
Kinetic energy is not conserved
hence, the collision is inelastic.