A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an ampl
itude of 0.101 mm and a frequency of 71.0 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x,t) = 
sin(kx + ωt), what are (a) , (b) k, and (c) ω, and (d) the correct choice of sign in front of ω?
1 answer:
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Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( )
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ ) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( )
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as
Here,
= Change in height
m = mass of super heroine
g = Acceleration due to gravity
The change in height will be,
The final position of the heroin is below the ground level,
The initial height will be the zero point of our system of reference,
Replacing all this values we have,
Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J