- 6.63×10-26 J. The energy of a photon is dictated by the equation E=h*c/λ, where h is plank’s constant and λ is the wavelength. Plank’s constant is 6.63×10-34 m2kg/s, and the frequency is 3 metres. The speed of light in a vacuum, c, is 3×108 metres per second. Put these all together to get 6.63×10-26 m2kg/s2. And a m2kg/s2 is a Joule of energy. Our answer is thus 6.63×10-26 Joules.
- 8.33 mJ. The energy of any particle with mass m0 and velocity v is shown in the Solution to Fascinating Frequencies 1. Plug 0.511 MeV and 1/3 c into the equation to get our answer. 8.33 microjoules.
- 2.214 m/s. Let t be the time the ball takes to fall. The average velocity of the ball is h/t. The terminal velocity is g*t. But the terminal velocity of the ball is twice the average since the initial is none. This tells us that t equals √(2h/g). And so the terminal velocity of a ball dropped from height h under gravity g is dictated by the equation 2h / √(2h/g). Plug in 1 metre and 9.807 m/s2 to get our answer. 2.214 metres per second.
- 1090 m/s. The force of gravity accelerates the ball downward at a rate of 9.807 m/s2. The radius of the ball is 10 centimetres and the density is 9.3×102 kg/m3. It’s volume is 1333π cm3, so it’s mass is 3.895×10-2 kilograms. The drag that will act on the ball is Cd*p*A*v2/2, where Cd is the drag coefficient, p is the density of the fluid, A is the frontal area, and v is the velocity relative to the fluid. The drag coefficient for a sphere is 0.47. The density of air is 1.23 kg/m3. And the frontal area of the ball is 100π cm2. Combine these to get a force of 0.009 v2kg/m. The velocity needed for this to equal 9.807 m/s2 is 1090 m/s. Therefore the terminal velocity of the ball is 1090 m/s.
- Trick question! it depends on the wind conditions. They can alter the density of air and change the time it takes to burn.
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