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julia-pushkina [17]
3 years ago
7

The potential difference in a simple circuit is 2 v and the resistance is 2 ω . what current flows in the circuit? answer in uni

ts of
a.
Physics
1 answer:
balandron [24]3 years ago
7 0
Hey

Potential Difference given is : 2V

Resistance is : 2 ohms

By Ohm's Law, one can easily utilize the relation :

v = ir

Where, { v , i , r } are the potential difference, current and Resistance Respectively.


Hence,
i =  \frac{v}{r}  =  \frac{2}{2}  = 1a

Hence, the Current is 1 Ampere
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A negatively charged rod is placed near two neutral metal spheres, as shown below. Which statements describe the charging method
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<h3>Answer;</h3>
  • <em>The spheres develop opposite charges. </em>
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  • <em>The spheres are charged through induction.</em>
<h3><u>Explanation;</u></h3>
  • <u><em>When a negatively charged rod is placed near two neutral metal spheres, the spheres will develop opposite charges, because the neutral metal spheres have both negative and positive charges. </em></u>From the basic law of electrostatics unlike charges attracts and like charges repel.
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Why does your boat float but not your ring?
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The answer to your question is C
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A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.
Ugo [173]

Answer:

(A) 10052.2 m/s²

(B)  0.00678 seconds

Explanation:

From the question,

(A) Applying

V² = U²+2as..................... Equation 1

Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.

make a the subject of the  equation

a = (V²-U²)/2s........................ Equation 2

Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

Substitute these values into equation 2

a = (68²-0²)/(2×0.230)

a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

make t the subject of the equation

t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

t = (68-0)/10052.2

t = 0.00678 seconds

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3 years ago
Consider the graph below that represents the variation of the velocity with respect to time of an object moving along the x - ax
vazorg [7]

In a velocity-time graph, the area under the curve represents the distance.

The distance traveled from 10s to 18 s is

\begin{gathered} d\text{ = }\frac{1}{2}\times10\times(18-10) \\ =\text{ 40 m} \end{gathered}

Final Answer: The distance traveled is 40 m from time 10 s to 18 s.

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