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nignag [31]
3 years ago
10

Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 2.0 104 kg (about 20 tons). The ro

cket engines are shut off, and you're coasting along with a constant velocity of ‹ 0, 28, 0 › km/s. As you pass the location ‹ 4, 4, 0 › km you briefly fire side thruster rockets, so that your spacecraft experiences a net force of ‹ 7.0 105, 0, 0 › N for 21 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?)
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

   r = <2.640 10⁶, 1.01 10⁸, 0> m

Explanation:

For this exercise we are going to solve it for each direction separately,

we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is

  r₀ = <4, 4, 0> 10³ m

X axis

the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off

They ask where it is after one hour t = 1 h = 3600 s

Let's apply Newton's second law

        Fₓ = m aₓ

        aₓ = Fₓ / m

        aₓ = 7.0 10⁵/2 10⁴

        aₓ = 3.5 10¹ m / s

Let's use kinematics to find the distance

for the first t₁ = 21 s the movement is accelerated

        x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²

        x₁ = x₀ + ½ aₓ t₁²

        x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5

        x₁ = 11717.5 m

this instant has a speed of

        vₓ = v₀ₓ + aₓ t

        vₓ = aₓ t ₁

        vₓ = 35  21

        vₓ = 735 m / s

the rest of the time there is no acceleration so it is uniform motion at this speed

        t₂ = 3600 - 21

        t₂ = 3576 s

         vₓ = x₂ / t₂

         x₂ = vₓ t₂

         x₂ = 735  3576

         x₂ = 2,628 10⁶ m

the total distance traveled in this direction is

         x_total = x₁ + x₂

         x_total = 11717.5 + 2.628 10⁶

         x_total = 2,640 10⁶ m

Y axis

on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of v_{oy} = 28 10³ m / s and there is no force on this axis F_{y} = 0

The movement in this axis is uniform,

       v_{y} = y / t

       y = v_{y} t

       y = 28 10³/3600

       y = 1.01 10⁸ m

the total distance is

       y_total = y₀ + y

       y_total = 4000 + 1.01 10⁸

       y_total = 1.01 10⁸ m

Z axis

the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0

the movement is uniform

         z = 0

the final position of the rocket after 1 h is

      r = <2.640 10⁶, 1.01 10⁸, 0> m

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Answer:

160 kg

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v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

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Mass of second car = 160 kg

Velocity of second car = 12 m/s

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a tire with inner volume of 0.0250m^3 is filled with air at a gauge pressure of 36.0 psi. If the tire valve is opened to the atm
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Answer: Escaped volume = 0.0612m^3

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According to Boyle's law

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P2 = atmospheric pressure= 14.696psi

V1 = volume of tire =0.025m^3

V2 = escaped volume + V1 ( since air still remain in the tire)

V2 = P1V1/P2

V2 = 50.696×0.025/14.696

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3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

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T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

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