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BartSMP [9]
2 years ago
11

Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal

Physics
2 answers:
Serhud [2]2 years ago
6 0

Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement

  • Force=87N
  • Displacement=4.5m

\boxed{\sf W=Fd}

  • W denotes to work done
  • F denotes to force
  • d denotes to displacement

\\ \sf\longmapsto Work\:Done=87(4.5)

\\ \sf\longmapsto Work\:done=391.5J

elena-s [515]2 years ago
5 0

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

  • W = F║d
  • where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

  • W = (87 N)(4.5 m)
  • W = 391.5 J  

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

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Answer:

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10 months ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
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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

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

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

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

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