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Finger [1]
2 years ago
15

Will the positioning of your Slinky along the east-west, north-south, or some other orientation affect your readings? Why?

Physics
1 answer:
Oxana [17]2 years ago
5 0

Positioning your Slinky along any direction different from its initial position will affect your reading, because there will be change in the magnetic field.

<h3>Effect of magnet on Slinky</h3>

If the Slinky is made of an iron alloy, it can be magnetized by itself. Moving the Slinky around can cause a change in the magnetic field, even if no current is flowing.

When there is a change in the magnetic field, the reading changes.

At any point, you change the orientation of the Slinky, you will need to zero the reading or adjust the Slinky back to its initial position, even if the sensor does not move.

Thus, Positioning your Slinky along any direction that is different to its initial position will affect your reading because there will be change in the magnetic field.

Learn more about magnetic field here: brainly.com/question/7802337

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Boyle's Law mainly involves _______.
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A machine does 500 j of work in 20 sec. What is the power of this machine?
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Explanation:

P=W/t

P=500/20

P=25 W

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What does the area between the line and the x axis represent on a velocity vs time graph
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the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period

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What kind of friction is occurring between a pencil and desktop when you flick the pencil?
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8 0
2 years ago
You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m
ki77a [65]

So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

\boxed{\sf{\bold{t = \frac{s}{v}}}}

With the following condition :

  • t = interval of the time (s)
  • s = shift or displacement (m)
  • v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

  • h = height or any other displacement at vertical line = 19.6 m
  • g = acceleration of the gravity = 9.8 m/s²
  • v = velocity = 343 m/s

What was asked :

  • \sf{\sum t} = ... s

Step by step :

  • Find the time when the object falls freely until it hits the water. Save value as \sf{\bold{t_1}}

\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}

\sf{t_1 = \sqrt{4}}

\sf{\bold{t_1 = 2 \: s}}

  • Find the time when the sound propagate through air. Save value as \sf{\bold{t_2}}

\sf{t_2 = \frac{h}{v}}

\sf{t_2 = \frac{19.6}{343}}

\sf{\bold{t_2 \approx 0.06 \: s}}

  • Find the total time \sf{\bold{\sum t}}

\sf{\sum t = t_1 + t_2}

\sf{\sum t \approx 2 + 0.06}

\boxed{\sf{\sum t \approx 2.06}}

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

3 0
2 years ago
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