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Tasya [4]
3 years ago
10

A runner starts at position zero and moves in one direction to position 12m 3. What is the distance traveled by the runner?

Physics
1 answer:
murzikaleks [220]3 years ago
5 0

<u>Distance = 12 m</u>

<u>Explanation:</u>

Distance covered by the runner is same as the total path that he has ran. Here, a person starts at 0 position and goes in one direction to position 12 m. Since the initial point is 0 and the final position is 12, the total distance is calculated by adding the 2 points as: 0+12 = 12 m. So the distance covered and the displacement both are same which is 12 meters.

Distance traveled is the full length of the path covered between two points. It is not a vector and the direction is nil and no negative sign.  Example : The distance walked by the teacher is 3.0 m ,The distance walked by a passenger is 5.0 m. Also, the distance traveled need not be equal to the magnitude of the displacement which is the distance between the two positions.

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If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.

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3 0
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A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t
vampirchik [111]

The new period is D) √2 T

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum 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>

\texttt{ }

\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}

T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}

T : T_2 = \sqrt{L} : \sqrt{2L}

T : T_2 = 1 : \sqrt{2}

\boxed {T_2 = \sqrt{2}\ T}

\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

3 0
3 years ago
Read 2 more answers
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