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: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
Answer:
a luminous ball of plasma
Answer:
1. The final velocity of the truck is 15 m/s
2. The distance travelled by the truck is 37.5 m
Explanation:
1. Determination of the final velocity
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Final velocity (v) =?
The final velocity of the truck can be obtained as follow:
v = u + at
v = 0 + (3 × 5)
v = 0 + 15
v = 15 m/s
Therefore, the final velocity of the truck is 15 m/s
2. Determination of the distance travelled
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Distance (s) =?
The distance travelled by the truck can be obtained as follow:
s = ut + ½at²
s = (0 × 5) + (½ × 3 × 5²)
s = 0 + (½ × 3 × 25)
s = 0 + 37.5
s = 37.5 m
Therefore, the distance travelled by the truck is 37.5 m