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zaharov [31]
3 years ago
12

Charge flows from low potential to high potential. O True or O False

Physics
1 answer:
inn [45]3 years ago
8 0

Answer:

False

Explanation:

Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).

Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

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If the kinetic energy of a particle is doubled, by what factor has its speed increased?
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a student attaches a 0.5 kg object to a 0.7 m string and rotates the object around her head and parallel to the ground. how much
Marina CMI [18]

The object would have a centripetal acceleration <em>a</em> of

<em>a</em> = (12 m/s)² / (0.7 m) ≈ 205.714 m/s²

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<em>T</em> = (0.5 kg) <em>a</em> ≈ 102.857 N ≈ 100 N

(rounding to 1 significant digit)

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3 years ago
Which does a reference point provide? Select two options. a position from which to measure future distance a set of standard uni
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Answer:

hope this helps you :)

Explanation:

An object is in motion if its position changes relative to another object. To decide if you are moving, you can use your chair as a reference point. A reference point is a place or object used for comparison to determine if something is in motion.

4 0
4 years ago
Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30
vekshin1

66.7 Watts

Explanation:

Let R_{1}=1.0 ohms, R_{2}=3.0ohms and R_{3}=6.0\:ohms. Since R_{2} and R_{3} are in parallel, their combined resistance R_{23} is given by

\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}

or

R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms

The total current flowing through the circuit <em>I</em> is given

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where

R_{Total}=R_{1}+R_{23}= 3.0\:ohms

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I=\dfrac{30\:V}{3.0\:ohms}=10\:A

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V_{1}=(10\:A)(1.0\:ohms)=10\:V

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P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W

4 0
3 years ago
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