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8090 [49]
3 years ago
8

what would you expect to happen to the acceleration if all friction were removed from the ramp, making the net force even higher

than 600n
Physics
2 answers:
Kruka [31]3 years ago
6 0

Answer:

If friction were removed from the ramp, the acceleration of the skateboarder will increase.

Explanation:

Gnoma [55]3 years ago
5 0

Answer:

Increase

Explanation:

The acceleration will increase more if all the friction were removed from ramp.

Acceleration is the rate of change of velocity with time.

Frictional force is a force that opposes motion of a body.

  • A large friction will reduce the velocity of a body significantly and bring it to stop, a state of no acceleration.
  • When friction is removed, the acceleration of a body will increase significantly since the velocity will change rapidly.
  • At each turn of time, the velocity keeps increasing due to absence of the inhibiting force of friction.
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Answer:

2) c) give-way vessel

3) a) With one short blast

Explanation:

2) A vessel that is required to take early substantial action to ensure avoiding  collision called Give way vessel

In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel

Therefore, the correct option is c) give-way vessel

3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast

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4 0
3 years ago
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
Crazy boy [7]

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

So

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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4 years ago
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