Answer:
Physical Properties of Sodium
Atomic number 11
Melting point 97.82°C (208.1°F)
Boiling point 881.4°C (1618°F)
Volume increase on melting 2.70%
Latent heat of fusion 27.0 cal/g
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Home Periodic table Elements Sodium
Sodium - Na
Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium
Atomic number
11
Atomic mass
22.98977 g.mol -1
Electronegativity according to Pauling
0.9
Density
0.97 g.cm -3 at 20 °C
Melting point
97.5 °C
Boiling point
883 °C
Vanderwaals radius
0.196 nm
Ionic radius
0.095 (+1) nm
Isotopes
3
Electronic shell
[Ne] 3s1
Energy of first ionisation
495.7 kJ.mol -1
You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'. In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.
Electrical power = (voltage) x (current) =
(6 volts) x (0.5 ampere) =
3 watts = 3 joules per second.
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
