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dedylja [7]
3 years ago
15

Earlier in this lesson, you spent some time thinking about events that involve matter and energy. Consider balls of different sh

apes and materials that reach different heights as they bounce off the same floor. What aspect of this event relates to matter? What aspect relates to energy? Font Sizes
Physics
1 answer:
Marianna [84]3 years ago
5 0

Answer:

See the answer below

Explanation:

T<em>he aspect of the event that relates to matter is </em><em>the materials and different shapes of the balls</em><em> while the aspect that relates to energy is t</em><em>he heights reached by the balls as they bounce off the same floor.</em>

Matter is generally defined as any substance that has weight and is able to occupy space. <u>Hence, the balls being made up of different materials and shapes means that they are made up of matter. </u>

Energy, however, is generally defined as the capacity to do work. The more the energy of a system, the more the work that can be done by the system. <u>The height reached by the balls is thus an indication of the amount of energy possessed by the balls.</u>

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Answer:

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2 years ago
What is the best use of an atomic model to explain the charge of the particles in thomson's beams
hoa [83]

The best use of an atomic model to explain the charge of the particles in Thomson's beams is:

<u>An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.</u>

<u>Explanation:</u>

In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.

Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

  • atoms are spheres of positive charge
  • electrons are dotted around inside

Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.

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3 years ago
We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth
Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = \frac{m}{m+M} \ v_o

for when n balls have collided

          v = \frac{m}{n \ m + M}  v₀

Now let's analyze the case of the bouncing ball (elastic)

     

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = \frac{m-M}{m + M}   v₀

we substitute in equation 1 to find v₂

            \frac{M}{m}  v₂ = v₀ -  \frac{m-M}{m+M}   v₀

            v₂ = \frac{m}{M}  ( 1 - \frac{m-M}{m+M} ) \ v_o

            v₂ = \frac{m}{M}  ( \frac{2M}{m+M} ) \ \ v_o

            v₂ = \frac{2m}{m +M}  \ v_o  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

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Serhud [2]

Answer:

b hope this helps

Explanation:

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The correct answer is decreases

The further away you are the weaker it would be. That's why at one point you stop being in the field and ti doesn't pull you towards it anymore. Proportionally, if you move towards the Earth then it increases. 
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3 years ago
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