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Colt1911 [192]
3 years ago
8

Draw a circuit that uses the resistors listed below as well as a 12V battery source. For your circuit, determine (a) the total r

esistance, (b) the total current, and (c) the total voltage over the entire system.
Physics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer: hello your questions lacks the required resistor values therefore i will provide a general answer using an example

answer : a) 14 ohms  b) 0.86 amps   c) 10.32 V

Explanation:

Assuming the resistors are : 3 ohms , 4 ohms and 5 ohms

Voltage source = 12V

<u>Assuming that the Resistors are in series </u>

<u>a) Determine Total resistance </u>

Req = R1 + R2 + R3

       = 3 + 4 + 5 = 14 ohms

<u>b) Total current </u>

Ieq = V / Req

      = 12 / 14 =  0.86 amps

<u>c) The Total Voltage over the entire system </u>

Vt = ∑ Voltage drops

    = ( 0.86 * 3 ) + ( 0.86 * 4 ) + ( 0.86 * 5 )

    = 10.32 V

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Answer:

The kinetic energy of the bullet is 5.4 × 10³ J

Explanation:

Hi there!

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

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Let´s convert the mass unit to kg so that our result is in Joules:

64 g · ( 1 kg / 1000 g) = 0.064 kg

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
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The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
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