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Mandarinka [93]
3 years ago
12

Question number 11 how did we found the answer ?

Physics
1 answer:
BaLLatris [955]3 years ago
6 0

Answer:

Option A. 57.14 Ω

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) = 100 Ω

Resistor 2 (R₂) = 400 Ω

Resistor 3 (R₃) = 200 Ω

Equivalent Resistor (Rₚ) =?

The equivalent resistor in the above circuit can be obtained as follow:

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

1/Rₚ = 1/100 + 1/400 + 1/200

Find the least common multiple (lcm) of 100, 400 and 200. The result is 400. Divide 400 by 100, 200 and 400 respectively and multiply the result with the numerator as shown

1/Rₚ = (4 + 1 + 2)/400

1/Rₚ = 7/400

Invert

Rₚ = 400/7

Rₚ = 57.14 Ω

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P = mv

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A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise.
Lera25 [3.4K]

The true statement about the CD is:

<h3><em>b. No net torque acts on it at all.</em></h3>

\texttt{ }

<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

\large {\boxed {a = \frac{ v^2 } { R } }

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

\texttt{ }

Centripetal Force can be formulated as follows:

\large {\boxed {F = m \frac{ v^2 } { R } }

<em>F = Centripetal Force ( m/s² )</em>

<em>m = mass of Particle ( kg )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

Let us now tackle the problem !

\texttt{ }

<em>Complete Question:</em>

<em>A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?</em>

<em>a. A net torque acts on it clockwise to keep it moving</em>

<em>b. No net torque acts on it at all.</em>

<em>c. A net torque acts on it counterclockwise to keep it moving</em>

<u>Given:</u>

angular velocity = ω = 5.0 revolutions per second

<u>Asked:</u>

net torque = Στ = ?

<u>Solution:</u>

Constant angular velocity → angular acceleration = α = 0 rad/s²

\Sigma \tau = I \alpha

\Sigma \tau = I (0)

\Sigma \tau = 0 \texttt{ Nm}

\texttt{ }

<h3>Conclusion:</h3>

The true statement about the CD is:

<em>b. No net torque acts on it at all.</em>

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

8 0
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Alkaline earth metals have a low density<br><br><br> true<br> false
Marta_Voda [28]

true

Explanation:

this is because melting point and boiling point decreases down the group because they are held together by attractions between positive nuclei and delocalised electrons

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