Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
<span>K.E = 0.5 * m * v^2 ( m = mass(Kg), V = Velocity(m/s)
= 0.5 * 8 * 5^2
= 4 * 25
= 100 J </span>
Answer
given,
For helium
Volume,V = 46 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₁ = ?
number of moles
we know
P V = n R T

n₁ = 1.89 moles
For oxygen
Volume,V = 12 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₂ = ?
number of moles
we know
P V = n R T

n₂ = 0.49 moles
Total volume of tank = 5 L
temperature of tank = 298 K
Partial pressure of helium


P₁ = 9.25 atm
Partial pressure of oxygen


P₂ = 2.39 atm
total pressure
P = P₁ + P₂
P = 9.25 + 2.39
P = 11.64 atm