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kirza4 [7]
1 year ago
10

The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of th

e planet X if the mass of planet X is 1.59 times that of Y, and its radius is 0.903 times the radius of Y.
Physics
1 answer:
serg [7]1 year ago
6 0

The escape velocity from the surface of the planet X is 2,249.2 m/s.

<h3>Escape velocity of planet X</h3>

v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}

where;

  • M is mass of the planet
  • r is radius of the planet
  • G is universal gravitation constant

\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 =  \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s

Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.

Learn more about escape velocity here: brainly.com/question/13726115

#SPJ1

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<h3>What is period of oscillation?</h3>

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<h3>Damping coefficient</h3>

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ln[-0.857 / cos(ω)] = -bx  

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<h3>at time, t = 2 cm, y = - 2cm</h3>

-2 = 3.5e^(-2bx) cos(2ω)

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solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

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ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

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