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jek_recluse [69]
3 years ago
5

Two rocks are moving through empty space in a straight line at the same speed. One has a mass of 10 kg, the other has a mass of

50 kg. Which statement about the rocks is true?
A.
Both will be stopped by the same amount of force acting against them.



B.
The one with less mass takes more force to stop.



C.
The one with greater mass takes more force to stop.



D.
Both will stop on their own without any force acting on them.
Physics
1 answer:
Yuki888 [10]3 years ago
6 0
A and D are definitely wrong because the two rocks have different masses, so it leaves us with B and C. The most logical answer to the question is
C <span>The one with greater mass takes more force to stop. 
</span>
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Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

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As the potential energy decreases kinetic energy increases.

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Two small space probes have been slowed to 10m/s as they approach the moon from the same direction. Probe 1 has a mass of 86kg a
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two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?
Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force F between two charged objects:

\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}},

where

  • k is coulomb's constant. k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2} in vacuum.
  • q_1 and q_2 are the signed charge of the objects.
  • r is the distance between the two objects.

For the two electrons:

  • q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}.
  • r = 1\times 10^{-10}\;\text{m}.
  • \displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}.

The sign of F is negative. In other words, the two electrons repel each other since the signs of their charges are the same.

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