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Sergio039 [100]
3 years ago
5

A 500 kg satellite is placed in a circular orbit 6394 km above the surface of the earth. At this elevation, the acceleration of

gravity is 4.09 m/s2. Knowing that its orbital speed is 20,000 km/h, determine the kinetic energy of the satellite. (Round the final answer to two decimal places.)
Physics
1 answer:
Digiron [165]3 years ago
6 0

Answer:

KE  = 7.7160 GJ

Explanation:

given data

mass = 500 kg

radius = 6394 km

acceleration of gravity = 4.09 m/s²

orbital speed = 20,000 km/h  = 20000 × \frac{1000}{3600}  = 5555.56 m/s

solution

we Kinetic energy is express as

KE = 0.5 × m × v²   .................1

here m is mass and v is velocity

put here value and we get

KE = 0.5 × 500 × 5555.56²

KE = 7716061728.400

KE  = 7.7160 GJ

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lana [24]

Answer:

Two scientists in a lab examining vials of urine they are analyzing for levels of excreted protein.

Explanation:

3 0
3 years ago
A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati
Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2

Initial acceleration of the rocket is 17.5 m/s²

v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

\Delta v=v_{e}\ln \frac{m_0}{m_f}

where,

m_0 = Initial mass of rocket with fuel

m_f = Final mass of rocket without fuel

v_e = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0
3 years ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
NemiM [27]

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

5 0
3 years ago
What is the source of energy for all tropical cyclones?
neonofarm [45]
The stored heat i guess..
3 0
3 years ago
A(n) 0.49 kg softball is pitched at a speed of 13 m/s. The batter hits it back directly at the pitcher at a speed of 23 m/s. The
garik1379 [7]
<h2>Impulse  = 4.9 kgm/s</h2>

Explanation:

Impulse is given by change of momentum.

Mass of softball = 0.49 kg

Initial velocity of softball = 13 m/s

Initial momentum = 0.49 x 13 = 6.37 kgm/s

Final velocity of softball = 23 m/s

Final momentum = 0.49 x 23 = 11.27 kgm/s

Change of momentum = Final momentum - Initial momentum

Change of momentum = 11.27 - 6.37

Change of momentum = 4.9 kgm/s

Impulse = Change of momentum

Impulse  = 4.9 kgm/s

4 0
3 years ago
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