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777dan777 [17]
3 years ago
7

Write a general scientific question that you will answer by doing this experiment.

Physics
1 answer:
il63 [147K]3 years ago
7 0

Answer:

How does newtons first two laws of motion apply to the toy car?

Explanation:

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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
3 years ago
Could magnets absorb electricity?
Phantasy [73]

If they are Conductive medals then yes.

They do attract or push away, cause sometimes they love each other or hate each other. x'D lol

5 0
3 years ago
Read 2 more answers
In a “minute to win it” game, cards are placed between cups to stack them. The contestant then pulls the card out in hopes that
Luden [163]

Answer:

There is no friction between the card and the cup.

Explanation:

4 0
3 years ago
Read 2 more answers
A dog runs 30 feet to the north then 5 feet to the south what is the displacement of the dog
Nuetrik [128]
To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.
5 0
3 years ago
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Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

4 0
3 years ago
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