The relationship between a cathode and an anode involves
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
Answer:
= 15.57 N
= 2.60 N
= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Explanation:
The weight of the sugar bag on Earth is:
g=9.81 m/s²
m=3.50 lb=1.59 kg
=m·g=1.59 kg×9.81 m/s²= 15.57 N
The weight of the sugar bag on the Moon is:
g=9.81 m/s²÷6= 1.635 m/s²
=m·g=1.59 kg× 1.635 m/s²= 2.60 N
The weight of the sugar bag on the Uranus is:
g=9.81 m/s²×1.09=10.69 m/s²
=m·g=1.59 kg×10.69 m/s²= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg