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 Virtual Laboratory is an interactive environment for creating and conducting simulated experiments: a playground for experimentation. It consists of domain-dependent simulation programs, experimental units called objects that encompass data files, tools that operate on these objects
Answer:
Differences
microscopic refers to substances visible to the naked eye
macroscopic are substances invisible to naked eye
Similarities
both refer to different scales that are useful to determining the size to different compounds.
Explanation:
Ima find more
Answer:
When the starting and ending points are the same, the total work is zero.
Explanation:
option ( D )correct
A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.
Answer:
450 kJ
Explanation:
Q = mCΔT
where Q is heat (energy),
m is mass,
C is specific heat capacity,
and ΔT is the temperature change.
Q = (1.2 kg) (4180 J/kg/°C) (100°C − 10°C)
Q = 451,440 J
Q ≈ 450 kJ
