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
Answer:
3.72 kJ
Explanation:
QH = 6.45 kJ
TH = 520 K
Tc = 300 K
Qc = ?
By use of Carnot's theorem
Qc / QH = Tc / TH
Qc / 6.45 = 300 / 520
Qc = 3.72 kJ
When you hold a spinning wheel, the wheel and you, chair included, form a system that obeys the principle of "conservation of angular momentum". This means that any changes in angular momentum within the system must accompanied by an equal and opposite change, so the net force is zero.
It is correct! good job :)
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration.
The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N.
a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r.
The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec.
So a_N = 114 m/sec^2.
g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
