Answer:
When two spheres, each with charge Q, are positioned a distance Rapart, they are attracted to ... doubled, the electric-force between the two spheres
Answer:
331.28 K
Explanation:
To solve this problem, you need to know that the heat that the water at 373 K is equal to the heat that the water at 285 K gains.
First, we will asume that at the end of this process there won't be any water left in gaseous state.
The heat that the steam (H20(g)) loses is equal to the heat lost because the change of phase plus the heat lost because of the decrease in temperature:
The specific Heat c of water at 298K is 4.18 kJ/K*kg.
The latent heat cl of water is equal to 2257 kJ/kg.
The heat that the cold water gains is equal to heat necessary to increase its temperature to its final value:
Remember that in equilibrium, the final temperature of both bodies of water will be equal.
Then:
Answer:
1.10134 * 10⁻⁹m⁻¹
Explanation:
K = 680Nm⁻¹
μ = ?
μ = (m₁ + m₂) / m₁m₂
compound = CO
C = 12.0 g/mol = 0.012kg/mol
O = 16.0g/mol = 0.016kg/mol
μ = (m₁ + m₂) / m₁m₂
μ = (0.012 + 0.016) / (0.012*0.016) = 145.83
v = 1/2πc * √(k/μ)
ν = 1/ 2*3.142* 3.0*10⁸ * √(630/145.83)
v = 5.30*10⁻¹⁰ * 2.078
v = 1.10134*10⁻⁹m⁻¹
The resultant<span> is the vector sum of 2 or more vectors. It is the conclusion of adding 2 or more vectors together. If </span>displacement <span>vectors A, B, and C are added together, the result will be vector R.</span>