a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube
1 answer:
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Answer:
Explanation:
Kinetic Energy formula:
KE = mv²
m=mass
v=speed
Given:
m=0.25kg
v=2.5m/s
Plug the values in:
KE = 1/2(0.25kg)(2.5m/s)²
KE = 0.78125 J (Joules)
Answer:
a) x_{cm} = m₂/ (m₁ + m₂) d , b) x_{cm} = 52.97 pm
Explanation:
The expression for the center of mass is
= 1 / M ∑
Where M is the total masses, mI and xi are the mass and position of each element of the system.
Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon
x_{cm} = 1 / (m₁ + m₂) (0+ m₂ x₂)
Let's reduce the magnitudes to the SI system
m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg
m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg
d = 128 pm = 128 10⁻¹² m
The equation for the center of mass is
x_{cm} = m₂/ (m₁ + m₂) d
b) let's calculate the value
x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷ 128 10-12
x_{cm} = 52.97 10⁻¹² m
x_{cm} = 52.97 pm