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Tasya [4]
3 years ago
6

A shopper walks eastward 3.2 meters and then westward

Physics
1 answer:
Musya8 [376]3 years ago
4 0
11.347 yards or 34.1 ft
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I need help. What is the correct answer A or B or C or D?
svet-max [94.6K]

Try grabbing a thin square material that would float normally. Try poking holes in it like shown in the examples. You will find your answer there! (Paper might work as long as the paper can handle water.)

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Why a steel ship becomes magnetized as it is constructed?
Inessa [10]
<em>Iron, and to a lesser degree, steel, can only become magnetised by passing an electrical current through it (an electromagnet). So a steel ship does not become magnetised in the accepted sense during construction. </em>

<span><em>However, any large mass of iron will affect the accuracy of a magnetic compass, causing it to deviate wildly from magnetic North. This problem was encountered when iron ships were first constructed in the mid-19 Century. It was overcome by mounting the compass in a 'binnacle', a housing containing two large soft iron balls either side of the compass itself, which counteracted the effect of the hull and balanced the compass so that it read correctly</em></span>
6 0
4 years ago
Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL &lt;&lt; L). Demuestre que su período aumenta en: ΔT = π
Kruka [31]

Answer:

 ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w = \sqrt{\frac{g}{L} }

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

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

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = 2\pi  \ \sqrt{\frac{L + \Delta L}{g} }

            T = 2\pi  \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }

in general the length increments are small ΔL/L «1, let's use a series expansion

           \sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...  

we keep the linear term, let's substitute

           T = 2\pi  \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L}  )  

if we do

           T = T₀ + ΔT

           

           T₀ + ΔT = 2\pi  \sqrt{\frac{\Delta L}{g} }  + \pi  \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}

            T₀ + ΔT = T₀ + \pi  \sqrt{\frac{1}{Lg} } \ \Delta L

            ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

4 0
3 years ago
A trapeze artist swings in simple harmonic motion with a period of 3.8 s.
zavuch27 [327]
As we know that time period of simple pendulum is given as

T = 2π √L/g

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Which insurance would contradict the big bang theory
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The Big Bang theory suggests that the universe is constantly expanding and that stars, galaxies and other entities are moving away from each other. If there were a galaxy moving closer to earth then that would contradict the Big Bang theory since the entities should be moving away from earth and from each other.
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