Answer:
3.71 eV
Explanation:
λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m
c = speed of electromagnetic wave = 3 x 10⁸ m/s
V₀ = stopping potential = 1.84 volts
W₀ = Work function of the metal = ?
Using the equation
= 5.94 x 10⁻¹⁹
= 3.71 eV
An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.
1 answer:
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