Explanation:
The object is moving along the parabola y = x² and is at the point (√2, 2). Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.
First, we need to find the radius of curvature. This is given by the equation:
R = [1 + (y')²]^(³/₂) / |y"|
y' = 2x and y" = 2:
R = [1 + (2x)²]^(³/₂) / |2|
R = (1 + 4x²)^(³/₂) / 2
At x = √2:
R = (1 + 4(√2)²)^(³/₂) / 2
R = (9)^(³/₂) / 2
R = 27 / 2
R = 13.5
So the centripetal force is:
F = m v² / r
F = m (5)² / 13.5
F = 1.85 m
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 answer to this question is a) sulfur
You look up the element in the periodic table. Subtract the atomic number (small number) from the mass number (big number). The answer is the number of neutrons.
Answer:0.00285714285 seconds
Explanation:
period=1 ➗ frequency
Period=1 ➗ 350
Period=0.00285714285 seconds