Physics, 09.01.2020 20:31 Nico8776

Suppose you are navigating a spacecraft far from other objects. the mass of the spacecraft is 2.5 × 104 kg (about 25 tons). the rocket engines are shut off, and you're coasting along with a constant velocity of < 0, 20, 0> km/s. as you pass the location < 6, 7, 0> km you briefly fire side thruster rockets, so that your spacecraft experiences a net force of < 3 × 105, 0, 0> n for 24.5 s. the ejected gases have a mass that is small compared to the mass of the spacecraft. you then continue coasting with the rocket engines turned off. where are you an hour later? (think about what approximations or simplifying assumptions you made in your analysis. also think about the choice of system: what are the surroundings that exert external forces on your system? )

Answers: 2

Show answers

Answers

Answer from: meaghan18

The final position of the spacecraft is $\vec r = \,[km]$ .

Let be a particle whose initial conditions are represented by vectors $\vec r_{o}$ , in meters and $\vec v_{o}$ , in meters per second. From statement we understand that acceleration ( $\vec a$ ), in meters per square second, is of the form:

$\vec a = \frac{1}{m}\cdot \vec F$ (1)

Where:

$\vec F$ - Vector force, in newtons.

- Mass, in kilograms.

Please notice that force is uniform in time.

By definitions of Position, Velocity and Differential Calculus, we have the following parametric equations for the velocity and position of the spacecraft in time:

$\vec v = \vec v_{o} + \frac{t}{m}\cdot \vec F$ (2)

$\vec r = \vec r_{o} + t\cdot \vec v_{o} +\frac{t^{2}}{2\cdot m}\cdot \vec F$ (3)

Where is the time, in seconds.

There are two stages in the motion of spacecraft:

1) Thruster rockets are turned on for 24.5 seconds. ( $\vec F \ne \vec O$ ).

2) Thruster rockets are turned off. ( $\vec F = \vec O$ ).

Stage 1:

( $\vec r_{o} = \,\left[\frac{m}{s} \right]$ , $\vec v_{o} = \,\left[\frac{m}{s} \right]$ , $\vec F = \,[N]$ , $t = 24.5\,s$ , $m = 2.5\times 10^{4}\,kg$ )

By (2):

$\vec v = + \left(\frac{24.5\,s}{2.5\times 10^{4}\,kg}\right)\cdot$

$\vec v = \,\left[\frac{m}{s}\right]$

By (1):

$\vec r = + (24.5)\cdot + \left[\frac{(24.5)^{2}}{2\cdot (2.5\times 10^{4})} \right]\cdot$

$\vec r = \,\left[m\right]$

Stage 2:

( $\vec r_{o} = \,\left[m\right]$ , $\vec v_{o} = \,\left[\frac{m}{s}\right]$ , $\vec F = \,\left[N \right]$ , $t = 3600\,s$ , $m = 2.5\times 10^{4}\,kg$ )

By (2):

$\vec v = \,\left[\frac{m}{s}\right]$

By (1):

$\vec r = + (3600)\cdot$

$\vec r = \,[m]$

$\vec r = \,[km]$

The final position of the spacecraft is $\vec r = \,[km]$ .

Link to a related question: link

Answer from: mahmudabiazp3ekot

The spacecraft is now at (1.1 × 10³, 7.2 × 10⁴, 0) km.

Explanation:

Assume that the thruster is the only source of force on the rocket. Acceleration in the first 24.5 seconds:

$\displaystyle a = \frac{\vec{F}}{m}=\frac{}{2.5\times 10^{4}} = \;\text{m}\cdot\text{s}^{-2} = \;\text{km}\cdot\text{s}^{-2}$ .

Distance travelled in the x-direction in the first 24.5 seconds:

$\displaystyle x = \frac{1}{2}\; a\cdot t^{2} = \frac{1}{2} \times 12\times (24.5)^{2} = 3.601\times 10^{3} \;\text{m} = 3.601\;\text{km}$ .

Distance travelled in the y-direction in the first 24.5 seconds:

$20\times 24.5 = 490\;\text{km}$ .

Velocity at the end of the first 24.5 seconds:

$v = u + a\cdot t = + 24.5\times =\;\text{km}\cdot\text{s}^{-1}$ .

Velocity of the spacecraft is constant between 24.5 seconds and 3600 seconds since there's no external force on the spacecraft. Displacement in that period of time:

$x = v\cdot t = (3600-24.5)\cdot = \;\text{km}$ .

Position vector of the spacecraft:

Initial: <6, 7, 0> km;After 24.5 seconds: <6, 7, 0> + <3.601, 490, 0> = <9.601, 497, 0> km;After 3600 seconds: <9.601, 497, 0> + <1051, 71510, 0> = <1060, 72007, 0> km.

Round to an appropriate number of significant figures. Position of the spacecraft after 3600 seconds: (1.1 ×10³, 7.2 × 10⁴, 0) km.

Answer from: Quest

lol

explanation:

Answer from: Quest

My answer is opptionc, just check with anyone else

Another question on Physics

Physics, 21.06.2019 20:00

Spend time observing or thinking about events that involve matter and energy. which events can you explain? which events can’t you explain?

Answers: 1

Answer

Physics, 21.06.2019 21:30

According to social exchange theory, altruistic behaviors benefit the individual who performs them.

Answers: 1

Answer

Physics, 22.06.2019 00:00

Did the proton move into a region of higher potential or lower potential? did the proton move into a region of higher potential or lower potential? because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.request answerpart bwhat was the potential difference that stopped the proton? express your answer with the appropriate units.î”v î” v = nothingnothingrequest answerpart cwhat was the initial kinetic energy of the proton, in electron volts? express your answer in electron volts.ki k i = nothing ev request answerprovide feedback

Answers: 2

Answer

Physics, 22.06.2019 13:10

Which of the following are used in an electrochemical cell? a. electrodes b. electrolyte c. terminals d. all of the above

Answers: 3

Answer

You know the right answer?

Suppose you are navigating a spacecraft far from other objects. the mass of the spacecraft is 2.5 ×...

Questions