Answer:
Change in momentum, ![\Delta p=6.3\ kg-m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3D6.3%5C%20kg-m%2Fs)
Explanation:
It is given that,
Mass of the basketball, m = 601 g = 0.601 kg
The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s
It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :
![\Delta p=mv\ cos\theta-(-mv\ cos\theta)](https://tex.z-dn.net/?f=%5CDelta%20p%3Dmv%5C%20cos%5Ctheta-%28-mv%5C%20cos%5Ctheta%29)
![\Delta p=2mv\ cos\theta](https://tex.z-dn.net/?f=%5CDelta%20p%3D2mv%5C%20cos%5Ctheta)
![\Delta p=2\times 0.601\times 6\times \ cos(29)](https://tex.z-dn.net/?f=%5CDelta%20p%3D2%5Ctimes%200.601%5Ctimes%206%5Ctimes%20%5C%20cos%2829%29)
![\Delta p=6.3\ kg-m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3D6.3%5C%20kg-m%2Fs)
So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.
Answer:
The number of photons per second that strike the given area is 2.668 x 10⁸ photons/second
Explanation:
Given;
intensity of the sunlight, I = 2.00 kJ·s−1·m^−2
area of incident, A = 5.2 cm² = 5.2 x 10⁻⁴ m²
Energy of incident photons per second on the given area;
E = IA
E = (2000)( 5.2 x 10⁻⁴)
E = 1.04 J/s
Energy of a photon is given is by;
![E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{(510*10^{-9})}\\\\E = 3.898*10^{-19} \ J/photon](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%20%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B%286.626%2A10%5E%7B-34%7D%29%283%2A10%5E8%29%7D%7B%28510%2A10%5E%7B-9%7D%29%7D%5C%5C%5C%5CE%20%3D%203.898%2A10%5E%7B-19%7D%20%5C%20J%2Fphoton)
The number of photons per second that strike the given area is;
![n = \frac{1.04 \ J/s}{3.898*10^{-19} \ J/photon} \\\\n = 2.668*10^{18} \ photons/second](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B1.04%20%5C%20J%2Fs%7D%7B3.898%2A10%5E%7B-19%7D%20%5C%20J%2Fphoton%7D%20%5C%5C%5C%5Cn%20%3D%202.668%2A10%5E%7B18%7D%20%5C%20photons%2Fsecond)
Therefore, the number of photons per second that strike the given area is 2.668 x 10⁸ photons/second
Answer:
The airfoil shape and wing size will both affect the amount of lift. The ratio of the wing span to the wing area also affects the amount of lift generated by a wing. ... The lift then depends on the velocity of the air and how the object is inclined to the flow. Air: Lift depends on the mass of the flow.
Please give brainliest!!!
Answer:
45° should be the correct one