Answer:air flow and the movement from where it’s coming from
Explanation:
Mass multiplied by acceleration produces force.
The acceleration is (v - 0)/t in this situation, where t seems to be the time it takes automobile A to come to a stop. According to Newton's third law of motion, the automobile produces this turning force of the wall, however the wall, which really is static and indestructible, forces an equal force back on the car.
According to Newton's third law, each action has an equal and opposite response. On this basis, you may deduce that a car driving into a wall would exert force on the wall. However, since the wall did not move, the automobile receives an equivalent force, causing it to collapse.
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Answer:
The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Explanation:
Given;
wavelength of ultraviolet light, λ = 270 nm
work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J
The energy of the ultraviolet light is given by;
![E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B%286.626%2A10%5E%7B-34%7D%20%29%283%2A10%5E%7B8%7D%29%20%7D%7B270%2A10%5E%7B-9%7D%20%7D%5C%5C%5C%5CE%20%3D%207.362%20%2A%2010%5E%7B-19%7D%20%5C%20J)
The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;
E = φ + K.E
K.E = E - φ
K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )
K.E = 3.677 x 10⁻¹⁹ J
K.E = ¹/₂mv²
mv² = 2K.E
velocity of the electron is given by;
![V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s](https://tex.z-dn.net/?f=V%20%3D%20%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%20%7D%5C%5C%5C%5CV%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%283.677%2A10%5E%7B-19%7D%29%20%7D%7B9.1%2A10%5E%7B-31%7D%20%7D%20%7D%5C%5C%5C%5CV%20%3D%208.99%2A10%5E%7B5%7D%20%20%5C%20m%2Fs)
the shortest de Broglie wavelength for the electrons is given by;
![\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bh%7D%7Bmv%7D%5C%5C%20%5C%5C%5Clambda%20%3D%20%5Cfrac%7B6.626%2A10%5E%7B-34%7D%20%7D%7B%289.1%2A10%5E%7B-31%7D%29%28%208.99%2A10%5E%7B5%7D%20%29%7D%5C%5C%5C%5C%5Clambda%20%3D%208.10%2A10%5E%7B-10%7D%20%5C%20m%5C%5C%5C%5C%5Clambda%20%3D%200.81%20%5C%20nm)
Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
(20 miles) x ( 1/45 hour/mile) =
(20/45) (hour) = <em>
4/9 hour = </em>26 minutes 40 seconds