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3 years ago

the plateau of tibet, the worlds largest plateau, is located which country , A,china- B, , north Korea -C,Japan- D,Mongolia

Geography

2 answers:

Reptile [31]3 years ago

4 0

A. China. The largest Plateau is called the Tibetan Plateau locates in China

Lena [83]3 years ago

3 0

A. China.----The Tibetan Plateau, also known in China as the Qinghai–Tibet Plateau or the Qing–Zang Plateau or Himalayan Plateau, is a vast elevated plateau in Central Asia and East Asia, covering most of the Tibet Autonomous Region and Qinghai in western China, as well as part of Ladakh in Jammu and Kashmir, India.

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2 years ago

A 215 kg rocket moving radially outward from Earth has a speed of 6.42 km/s when its engine shuts off 307 km above Earth's surfa

Vinvika [58]

<h2>GIVEN DATA</h2>

Mass of Rocket = m = 215 kg

Mass of Earth = M = $5.97*10^{24}$ kg

Radius of Earth = R = $6.37*10^6$ m

Gravitational Constant = G = $6.67*10^-11 \;\;m^3kg^{-1}s^{-2}$

Speed of Rocket = 6.42 km/s

Initial Height of Rocket from Earth's Surface = 307 km = $3.07*10^5$ m

Final Height of Rocket from Earth's Surface = 731 km = $7.31*10^5$ m

Initial Height from Earth's Centre = $R_i$ = $3.07*10^5 + 6.37*10^6 = 6.677*10^6$ m

Final Height from Earth's Surface = $R_f$ = $7.31*10^5 + 6.37*10^6 = 7.101*10^6$ m

(a) Kinetic Energy at Final Height = $K.E_f$

(b) Maximum Height of Rocket above Earth's Surface = $H_{max}$

<h2>EXPLANATION</h2>

Part (a):

As drag air is negligible, energy will be conserved.

$\therefore$ ΔE = 0

$K.E_i + U_i = K.E_f + U_f$

$K.E_f = U_f - K.E_i - U_i$

where, U is the potential energy of the system.

$K.E_f=\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}\;\;\;\;----------\;(1)$

Substituting Values and simplifying,

$K.E_f = 3.665*10^9 J$

Part (b):

The rocket will come to rest after reaching the maximum height. Therefore, its final velocity and consequently final kinetic energy will be zero.

$i.e.\;\;v_f = 0\;\;\;\&\;\;\; K.E_f=0$

Equation (1) will become,

$0\;=\;\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}$

$or\;\;\frac{1}{2}v_i^2= GM(\frac{1}{R_i}-\frac{1}{R_f})\\\\\frac{1}{2GM}v_i^2= (\frac{1}{R_i}-\frac{1}{R_f})\\\\\therefore\; R_f = \frac{2GMR_i}{2GM-v_i^2R_i}$

Substituting values and simplifying,

$R_f = 10.20*10^6 m$

which is the distance from the Earth's centre. To find the height of rocket from Earth's surface, we simply subtract the Earth's radius from above result.

$H_{max} = R_f - R$