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Bezzdna [24]
4 years ago
5

Ryan wants to transform electrical energy into light energy. He decides to build a complete circuit that will operate a light bu

lb. What materials does he need? A) a battery, wire, plastic clips, and a bulb B) a battery, string, plastic clips, and a fan C) a source of electrical energy, wire, metal clips, and a bulb D) a source of electrical energy, string, metal clips, and a bulb
Physics
2 answers:
alex41 [277]4 years ago
7 0

Answer:

C) a source of electrical energy, wire, metal clips, and a bulb

Explanation:

Since Ryan want to convert electrical energy to light energy so here we need a device which will give us electrical energy and also we need a source of light.

So here here in this experiment we need to convert this electrical energy into light energy so we need to connect the electrical bulb with the source of electrical energy so that the bulb will glow and give us light energy

Now here in order to connect the bulb with the battery we need connecting wires and metal clips to connect bulb with the battery or electrical energy source.

so here correct answer is

C) a source of electrical energy, wire, metal clips, and a bulb

Fofino [41]4 years ago
5 0
C. is the answer. You can't use plastic because it does not conduct electricity and string does not conduct electricity either so the circuit will end before it get to the bulb. 
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Which of the following is best described as a synthesis reaction? (2 points) 2HgO → 2Hg + O2 AgNO3 + NaCl → AgCl + NaNO3 HCl + N
Shtirlitz [24]

2Na + Cl₂ → 2NaCl

Explanation:

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4 years ago
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Afina-wow [57]
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3 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

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Explanation:

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