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1 year ago

Which types of questions can most likely be answered through a scientific investigation?

1 answer:

4 0

Subjective questions can be answered most likely through well-designed scientific investigations. Investigating peoples opinions or favorites does not involve any scientific investigations.

<h3>What are scientific investigation?</h3>

Scientific investigations include study on different natural phenomenon through well designed research methodologies. They include a preliminary hypothesis based on observations and various scientific records helps for reference.

Through well designed scientific experiments we can solve subjective questions and can be helpful in solving different environmental or social issues.

Peoples opinion or objective types questions are not targeted at scientific investigation, hence, option b is correct.

To find more on scientific investigations, refer here:

brainly.com/question/8386821

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The vertical component illustrates ____ motion and the horizontal velocity component ___.

Bond [772]

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3 years ago

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>

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<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>

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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 !

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<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

$\Sigma \tau = I \alpha$

$\Sigma \tau = I (0)$

$\Sigma \tau = 0 \texttt{ Nm}$

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<h3>Conclusion:</h3>

The true statement about the CD is:

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

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<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

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

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

#LearnWithBrainly

8 0

3 years ago

Find the magnitude of the gravitational force a 63.5 kg person would experience while standing on the surface of Earth with a ma

Anastaziya [24]

Answer: 3976N

Explanation:

Using the formula for calculating gravitational force between two masses, we have

F = GMm/r^2

Where G is the gravitational constant

M and m are the masses

r is the distance between the masses

F= 6.673 × 10-¹¹ × 5.98 × 10²⁴ × 63.5/ (6.37 × 10^6)^2

F= 2.533×10^16/6.37×10^12

F= 0.3976×10⁴N

F= 3976N

3 0

3 years ago

A girl standing upright exerts a pressure of 15000 N/m2 on the floor. Given that the total area of contact of shoes and the floo

lapo4ka [179]

Explanation:

F=15000/0.02

=300N

total area contact of the shoes and floor when stood in one foot=0.02m^2/2=0.01m^2

P=F/A

=300/0.01

=30 000Pa/30 000Nm^-2

in another way=15000×2

=30000N/m^2

5 0

3 years ago

A vertical spring (ignore its mass), whose spring stiffness constant is 880 n/m, is attached to a table and is compressed down 0

sladkih [1.3K]

When the spring is compressed, the potential energy stored in the spring is:
$U= \frac{1}{2}kx^2$
where k=880 N/m is the spring constant and $x=0.170 m$ is the compression of the spring. By using these numbers, we get
$U= \frac{1}{2}(880 N/m)(0.170 m)^2=12.7 J$

When the spring is released with the ball over it, all the potential energy is converted into kinetic energy of the ball:
$U=K= \frac{1}{2} mv^2$
So by using m=0.300 kg, and re-arranging the formula, we can calculate the velocity of the ball:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 12.7 J}{0.300 kg} }=9.2 m/s$

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3 years ago