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Illusion [34]
2 years ago
4

An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his back has a mass of 146 k

g. The astronaut begins a space walk at rest, with a completely filled propulsion unit. During the space walk, the unit ejects some gas with a velocity of 32 m/s. As a result, the astronaut recoils with a velocity of 0.39 m/s. After the gas is ejected, the mass of the astronaut (now wearing a partially empty propulsion unit) is 165 kg. What percentage of the gas was ejected from the completely filled propulsion unit?
Physics
1 answer:
Ksenya-84 [330]2 years ago
4 0

Answer:

10.96%

Explanation:

Mass of astronaut and empty tank

146 kg

Mass of astronaut and partially empty tank

162 kg

Mass of gas in the tank

162-146 = 16 kg

Momentum after gas release

m_1v_1+m_2v_2=0\\\Rightarrow m_1=-\frac{m_2v_2}{v_1}\\\Rightarrow m_1=-\frac{-162\times 0.39}{32}\\\Rightarrow m_1=1.97\ kg

Total mass of gas = 1.97+16 = 17.97 kg

Percentage depleted

\frac{1.97}{17.97}\times 100=10.96\%

The percentage of gas ejected is 10.96%

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yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
2 years ago
Please help!!!!
murzikaleks [220]

A simple electromagnet consisting of a coil of wire wrapped around an iron core. <u><em>A core of ferromagnetic material like iron serves to increase the magnetic field created.</em></u> The strength of magnetic field generated is proportional to the amount of current through the winding.

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5 0
2 years ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
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A solid spherical insulator has radius r = 2.5 cm, and carries a total positive charge q = 8 × 10-10 c distributed uniformly thr
ira [324]
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5 0
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Two equal length of wire made of the same material but of different diameters have an effective resistance of 0.8 ohm when they
nata0808 [166]

Answer:

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Explanation:

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