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AnnZ [28]
3 years ago
13

A rocket takes off from Earth and reaches a speed of 105 m/s in 18.0 s. If the exhaust speed is 1,200 m/sand the mass of fuel bu

rned is 110 kg, what was the initial mass (in kg, including the initial fuel) of the rocket?
Physics
1 answer:
Amanda [17]3 years ago
7 0

Answer:

The initial mass of the rocket is 526.2 kg.

Explanation:

Given that,

Final speed = 105 m/s

Time = 18.0 s

Exhaust initial speed = 1200 m/s

Mass of burned fuel = 110 kg

We need to calculate the initial mass

The velocity change of rocket under gravity is defined as,

v=u\ ln(\dfrac{m_{i}}{m})-gt....(I)

We know that,

m=m_{i}-m_{bf}

Put the value of m in equation (I)

v=u\ ln(\dfrac{m_{i}}{m_{i}-m_{bf}})-gt

m_{i}=\dfrac{m_{bf}}{1-e^-{\dfrac{v+gt}{u}}}

m_{i}=\dfrac{110}{1- e^-{\frac{105+9.8\times18}{1200}}}

m_{i}=526.2 kg

Hence, The initial mass of the rocket is 526.2 kg.

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Answer:

Δθ = 15747.37 rad.

Explanation:

  • The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
  • Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

       \omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)

  • Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

       \omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)

  • Solving for Δθ in (2):

       \theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)

  • The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

       \theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)

  • Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

       \omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)

  • Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      \theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)

  • The total angular displacement is just the sum of (3), (4) and (6):
  • Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
  • ⇒ Δθ = 15747.37 rad.
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