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hjlf
3 years ago
8

By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a

re you answering by doing this experiment?
Physics
2 answers:
NemiM [27]3 years ago
8 0

Answer:

Sample Response: How do changes in voltage or resistance affect current in an electric circuit?

Explanation:

andrezito [222]3 years ago
5 0

Broad questions that we can answer by doing this experiment are <em><u>the effects of electric current</u></em>

<h3>Further explanation</h3>

Electric current is the amount of electric charge that flows each unit of time

Electric current occurs due to the movement of electrons due to the difference in potential or voltage (from high potential to low potential) between two points

Electrons will flow through the conducting wire that functions as a conductor

Ohm's Law states that

<em>The potential difference is proportional to the electric current as long as the resistance is constant</em>

\displaystyle I=\frac{V}{R}\\\\V=I\times R

A simple electrical circuit consists of a voltage source (battery) and a lamp

Ampermeters to measure the strength of the current, must be installed in series with the load to be measured

By changing the voltage source, with constant resistance from a conductor, different current measurements will be obtained. The greater the voltage, the greater the resulting current.

<h3>Learn more</h3>

electrons flow through the device

brainly.com/question/4438943

Keywords : basic electric circuits, Ohm’s law, experiment

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Masja [62]

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = \sqrt{3^{2} +2^{2}  } = \sqrt{13} = 3.61cm = 0.036m

r₂ = \sqrt{4^{2} + 3^{2}  } = \sqrt{25} = 5cm = 0.05m

electric potential V = \frac{kq}{r}

change in potential ΔV = V_{1} - V_{2}

ΔV = \frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} } , where q_{1} = q_{2}=2.00μC

ΔV = 2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × (\frac{1}{0.036} - \frac{1}{0.05} )

ΔV= 2.789×10⁵

\frac{1}{2}mv^{2} = ΔV × q₃

\frac{1}{2} ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

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3 years ago
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Eduardwww [97]

Answer:

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freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

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Explanation:

Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.

Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.

On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.

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Answer:

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