If the wavelength of an electromagnetic wave is too long or too short,

Physics

2 answers:

stealth61 [152]3 years ago

4 0

Answer:

it is no longer a wave

Explanation:

I know because i did it

Evgesh-ka [11]3 years ago

3 0

it is no longer and electromagnetic wave.

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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo

makkiz [27]

Answer:

(a) The speed of the target proton after the collision is: $V_{2f} =433(m/s)$ , and (b) the speed of the projectile proton after the collision is: $v_{1f}=250(m/s)$ .

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle: $m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}$ , and y axle: $0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}$ . Now replacing the value given as: $v_{1i}=500(m/s)$ , $\beta_{1}=+60^{o}$ for the projectile proton and according to the problem $\beta_{1}and\beta_{2}$ are perpendicular so $\beta_{2}=-30^{o}$ , and assuming that $m_{1}=m_{2}$ , we get for x axle: $500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}$ and y axle: $0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}$ , then solving for $v_{2f}$ , we get: $v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}$ and replacing at the first equation we get: $500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}$ , now solving for $v_{1f}$ , we can find the speed of the projectile proton after the collision as: $v_{1f}=250(m/s)$ and $v_{2f}=\sqrt{3}*v_{1f}=433(m/s)$ , that is the speed of the target proton after the collision.