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wariber [46]
2 years ago
11

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during

the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.
What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?

A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2
Physics
1 answer:
Katarina [22]2 years ago
3 0

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

  • From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
  • From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

  • The speed increases from 0 m/s to 1341 m/s.
  • The time elpased is 2.0 min.
  • 1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s}  = 11 m/s^{2}

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

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The differential equation below models the temperature of an 88°C cup of coffee in a 24°C room, where it is known that the coffe
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Answer:

y = 24+64\cdot e^{-150\cdot t}

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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl
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Hi there!

a)

We can begin by using the equation for energy density.

U = \frac{1}{2}\epsilon_0 E^2

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}

Creating a Gaussian surface being the lateral surface area of a cylinder:
A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}

Now, we can calculate the energy density using the equation:
U = \frac{1}{2} \epsilon_0 E^2

Plug in the expression for the electric field and solve.

U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
V = \pi r^2L \\\\dV = 2\pi rLdr

Now, we can take the integral of the above expression. Let:
r_i = inner cylinder radius

r_o = outer cylindrical shell inner radius

Total energy-field energy:

U = \int\limits^{r_o}_{r_i} {U_D} \, dV =   \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr

Plug in the equation for the electric field energy density and solve.

U =   \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\

Bring constants in front and integrate. Recall the following integration rule:
\int {\frac{1}{x}} \, dx  = ln(x) + C

Now, we can solve!

U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}

And here's our equation!

3 0
2 years ago
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