What is the difference between brake horsepower and thrust horsepower?
1 answer:
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Answer:
mass consumed by 235U each day = 2 kg
Explanation:
electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV
= 6.24151× 10²¹ MeV/s
thermal energy = 0.420 * 250 = 105 MeV
= 5.94 × 10¹⁹ fission/second
=5.94 × 10¹⁹× 24 × 60 ×60)
= 5.13 × 10²⁴ fission/day
mu = 235.04393 × 1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg
M = mu ×5.13 × 10²⁴
= 390.1729× 10⁻²⁷ ×5.13 × 10²⁴
M = 2 kg(approx.)
mass consumed by 235U each day = 2 kg
Answer:
(a)
(b)
Explanation:
Hello.
(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:
(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:
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Answer:
The speed of the student just before she lands, v₂ is approximately 8.225 m/s
Explanation:
The given parameters are;
The mass of the physic student, m = 43.0 kg
The height at which the student is standing, h = 12.0 m
The radius of the wheel, r = 0.300 m
The moment of inertia of the wheel, I = 9.60 kg·m²
The initial potential energy of the female student, P.E.₁ = m·g·h₁
Where;
m = 43.0 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = 12.0 h
∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J
The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;
The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0
∴ K₁ = 0 + 0 = 0
The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0
Where;
h₂ = 0 m
The final kinetic energy, K₂, of the wheel and student is give as follows;
Where;
v₂ = The speed of the student just before she lands
ω₂ = The angular velocity of the wheel just before she lands
By the conservation of energy, we have;
P.E.₁ + K₁ = P.E.₂ + K₂
∴ m·g·h₁ + = m·g·h₂ +
Where;
ω₂ = v₂/r
∴ 5061.96 J + 0 = 0 +
5,061.96 J = 21.5 kg × v₂² + 53. kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²
v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²
v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s
The speed of the student just before she lands, v₂ ≈ 8.225 m/s.