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Natalija [7]
2 years ago
6

Following Newton's Third Law, when a cannon goes off the cannon ball exerts a force on the cannon and the cannon exerts the same

force back on the cannon ball. Why does the cannon ball travel severl hunder feet while the cannon itself only moves a few inches?
Question 5 options:

Because the mass of the cannon ball is much less than the cannon

Because the cannon is so much larger in size than the cannon ball

Because the cannon ball is more aerodynamic than the cannon

Because the forces were not really equal
Physics
1 answer:
bagirrra123 [75]2 years ago
6 0

Answer:

  • <u>First choice:</u><u><em> Because the mass of the cannon ball is much less than the cannon</em></u>

Explanation:

Indeed, <em>Newton's Third Law</em>, i.e. the action-reaction law, states that any action (force) will have a reaction (force) of same magnitude but opposite direction.

That means that when a cannon goes off the cannon ball exerts a force on the cannon and the cannon exerts the same force back on the cannon ball.

To find out how much the cannon ball and the cannon itsel move, you must consider Newton's second law.

  • F = m×a (force equal mass times acceleration).

Clearing the acceleration you get:

  • a = F / m

Then, since the mass is in the denominator and both the force that the cannon ball exerts on the cannon and the cannon exerts on the cannon ball are equal in magnitude, then the body that has the smaller mass (the cannon ball)  will experience a greater acceleration, which is stated by the first choice: because the mass of the cannon ball is much less than the cannon.

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Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

T = \frac{1}{f}

Where:

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T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

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Therefore, the amplitude is 10 cm.          

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Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

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Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

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Also;

F = \frac{MV}{t}

solving this two equations together;

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m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

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substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

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