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3 years ago

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If you could travel to a star 25 light-years away and return to earth at nearly the speed of light, how much time would elapse o

n earth?

2 answers:

anastassius [24]3 years ago

7 0

Millions and millions of years

lyudmila [28]3 years ago

6 0

50 years, would it not be, because time moves on no matter what, so 25ish plus 25ish would be 50 years. Good question

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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