the resistivity of a given wire of cross session area 0.7mm² is 49×10⁴.calcuate the resistance of a 2m lenght of wire
1 answer:
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Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w = w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) ( w₀)²
Kf = ½ w₀²
we look for the relationship between the kinetic energy
=
K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision