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Anettt [7]
3 years ago
9

An 82.0 kg spacewalking astronaut pushes off a 655 kg satellite, exerting a 95.0 N force for the 0.530 s it takes him to straigh

ten his arms. How far apart are the astronaut and the satellite after 1.0 min.?
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

41.41 m

Explanation:

When force F is applied on an object of mass m for time t and velocity v₁ is created

F X t = mv₁

F = 95 N , t = .53 s, m = 655 kg

95 x .53 = 655 x v₁

v₁ = .0768 m/s

Applying conservation of momentum on man and satellite

m₁ v₁ = m₂v₂

655 x .0768 = 82 xv₂

v₂ = .6134 m/s

their relative velocity

= .6134 + .0768

= .6902 ( they are in opposite direction )

After 60 second distance between them

= 60 x .6902 m

= 41.41 m

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he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee
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Answer:

a

 F =  -1.07 *10^{-8} \  N

b

F_r  =  1.07 *10^{-8} \  N

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

F =  \frac{k *  [Q_{Li}  ] * [Q_{O}  ]  }{ r^2}

Here k is known as the proportionality constant with value k = 2.31  *  10^ {-28} J \cdot m

substituting -2 for Q_{O} i.e the charge on oxygen , +1 for Q_{Li} i.e the charge on Lithium and [0.140 + 0.068 ] nm= 0.208 nm =  0.208*10^{-9} for r

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F =  \frac{ 2.31  *  10^ {-28}*  +1   * -2   }{ ( 0.208*10^{-9} )^2   }

F =  -1.07 *10^{-8} \  N

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

F_r  =  1.07 *10^{-8} \  N

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