<span>The three types of seismic waves produced by an earthquake are primary, secondary, and (D.) surface.</span>
Answer:
Explanation:
A 40kg child throw stone of 0.5kg
At a direction of 5m/s
Recoil can be calculated using recoil of a gun formula
m_1•v_1 + m_2•v_2
m_1•v_1 = -m_2•v_2
The negative sign show that the momentum of the boy is directed oppositely to that of the stone
m_1 Is mass of boy
v_1 is the recoil velocity of the boy
m_2 is mass of stone
v_2 is the velocity of stone
Then,
m_1•v_1 = -m_2•v_2
40•v_1 = -0.5 × 5
40•v_1 = -2.5
v_1 = -2.5 / 40
v_1 = -0.0625 m/s
The recoil velocity of the boy is 0.0625 m/s
Answer:
X: Always attractive
Y: Infinite range
Z: Attractive or repulsive
ANSWER IS C
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