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konstantin123 [22]
2 years ago
13

Explain why the rocket can move at constant speed in outer space.

Physics
2 answers:
iren [92.7K]2 years ago
4 0

Answer:

When the rocket's engines are fired up, the force of gravity is disturbed, and the rocket takes off in the air. Later, as the rocket's fuel runs out, it slows down, comes to a halt, and eventually plummets to Earth's surface. Forces can affect objects in space as well. Spaceships are constantly in motion while traversing the solar system.

Explanation:

Please read!

disa [49]2 years ago
3 0

Answer:

Explanation:

The rocket works because of the law of conservation of linear momentum. The law of conservation of linear momentum is very important in physics. Momentum is defined as the mass of an object times its velocity.

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The lowest energy of electron is the ground state.

Explanation:

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Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres
inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

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Given, v_{1} =1.4 m/s

Now from equation continuity,

v_{1} A_{1} = v_{2} A_{2}.

Here, A_{1} and A_{2} are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}.

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}

v_{2} = 4 \times 1.4 = 5.6 m/s

Substituting these values  with the density of water is 1000 \ kg/m^3 in pressure difference formula we get.

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa

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For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem: 

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<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

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(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

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<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

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0=(8.549m/s)+(-9.81m/s^2)(t) 

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s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m</span>

 

Answer:

<span>1.925m</span>

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