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scoundrel [369]
2 years ago
9

:)

Physics
1 answer:
sweet-ann [11.9K]2 years ago
8 0

Answer:

D

Explanation:

1.) The reaction is at dynamic equilibrium.

A: Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

2.) Which statement about the reaction is necessarily correct?

A: Both calcium carbonate and sodium carbonate are being produced.

3.) Both calcium carbonate and sodium carbonate are being produced.

A: The reaction is reversible.

4.) What is the fastest motion that can be measured in any frame of reference?

A: 300,000 km/s

5.) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north.

Why are we able to use these motions to describe the motion relative to the ground?

A: The people are moving much slower than the speed of light so the ground acts as a frame of reference.

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An object is attached to a vertical spring and bobs up and down between points a and
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 it would be at either A or B. 
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Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

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3 years ago
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Which factor MOST directly affects the flow of ocean currents?
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Read 2 more answers
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
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