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zhannawk [14.2K]
3 years ago
11

The helicopter was deformed and destroyed in the __?___ collision.

Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

The helicopter was deformed and destroyed in the inelastic collision.

Explanation:

  • When two object collide there exist two way of colliding: elastic collision and inelastic collision.
  • Two terms are considered during the collision: kinetic energy and momentum.
  • If both of these terms are conserved in any collision then there is no significant loss of property, this is called as elastic collision.
  • If only momentum is conserved but kinetic energy is converted into other forms then it is inelastic collision. In inelastic collision, the energy is lost in the form of vibration, sound etc. causing the damage to colliding object.
  • Hence the deformation of helicopter was due to inelastic collision.
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What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?
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3 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

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E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

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Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

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So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

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I think the answer is c
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