A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s
1 answer:
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The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.
Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.
When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.
Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.
When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
Answer:
λ = 396.7 nm
Explanation:
For this exercise we use the diffraction ratio of a grating
d sin θ = m λ
in general the networks works in the first order m = 1
we can use trigonometry, remembering that in diffraction experiments the angles are small
tan θ = y / L
tan θ = = sin θ
sin θ = y / L
we substitute
= m λ
with the initial data we look for the distance between the lines
d =
d = 1 656 10⁻⁹ 1.00 / 0.600
d = 1.09 10⁻⁶ m
for the unknown lamp we look for the wavelength
λ = d y / L m
λ = 1.09 10⁻⁶ 0.364 / 1.00 1
λ = 3.9676 10⁻⁷ m
λ = 3.967 10⁻⁷ m
we reduce nm
λ = 396.7 nm