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trapecia [35]
2 years ago
11

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed

v0(v < vesc). Express in terms of v0, rE, ME, and G.
Physics
1 answer:
olga55 [171]2 years ago
7 0

Initially, the energies are:

U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\&#10;=K_{i}=\frac{1}{2} m v_{0}^{2}

At final point, the energies are:

U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\&#10;K_{f}=\frac{1}{2} m(0)^{2}=0

Using conservation law of energy,

-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\&#10;-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\&#10;\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\&#10;\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}

The equation is further simplified as:

r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\&#10;h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\&#10;&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\&#10;& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}

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Shkiper50 [21]
A region around a charged particle or object within which a force would be exerted on other charged particles or objects
5 0
3 years ago
6000 kg train moving 5 m/sec to east collides with 5000 kg train moving 3 m/sec to west. What is their velocity
Svetlanka [38]

The final velocity is 1.37 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v

where:

m_1 = 6000 kg is the mass of the first train

u_1 = 5 m/s is the initial velocity of the first train (we take east as positive direction)

m_2 = 5000 kg is the mass of the second train

u_2 = -3 m/s is the initial velocity of the second train

v is the final combined velocity of the two trains

Re-arranging the equation and substituting the values, we find:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(5)+(5000)(-3)}{6000+5000}=1.37 m/s

And the positive sign indicates their final direction is east.

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

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brainly.com/question/9484203

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8 0
3 years ago
Write fulk form of FPS?<br>​
zzz [600]

Answer:

<u>Foot per second. Foot-pound-second system. Frames per second, the frequency (rate) at which consecutive images (frames) appear on a display.</u>

Explanation:

:)

4 0
2 years ago
a ball thrown vertically downwatd stikes a horizontal surface with a speed of 15 meters per seconds if then bounces and reaches
Anna71 [15]

but why was it thrown is the real question

3 0
2 years ago
An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then
gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

a=\frac{dv}{dt}=1.76

So, the velocity can be obtained by integrating this expression:

v=1.76t

The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

At this time the velocity is: v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}

This velocity remains constant in the section 2, so for that section the movement equation is:

x=v*t\\t=\frac{x}{v}

The left distance is 89 meters, and the velocity is 6.2225\frac{m}{s}, so:

t=\frac{89}{6.2225}=14.303s

So, the total time is 14.303+3.5355s=17.838s

7 0
2 years ago
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