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Alona [7]
3 years ago
15

A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object o

f mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.
Physics
1 answer:
lara [203]3 years ago
8 0

Answer:

the answers the correct one is cη

Explanation:

In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so

              sin θ = θ

 

with this approach the equation will be surveyed

     d² θ / dt² = - g / L sin θ

It is reduced to

      d² θ / dt² = - g / L θ

in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%

The angle is related to the height of the pendulum

         sin θ = h / L

         h = L sin θ.

Therefore the student must be careful that the height is small.

When reviewing the answers the correct one is cη

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A coffee filter of mass 1.2 grams dropped from a height of 1 m reaches the ground with a speed of 0.8 m/s. How much kinetic ener
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<u>Explanation:</u>

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Here, the potential energy of the coffee filter is getting converted into kinetic energy of the coffee filter and some energy is lost by it which is gained by the air molecules in the form of kinetic energy.

So, calculating the potential energy of coffee filter, we use the equation:

P = mgh

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m = mass of coffee filter = 1.2 g = 1.2\times 10^{-3}kg    (Conversion used: 1 kg = 1000 g)

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h = height of coffee filter = 1 m

Putting values in above equation, we get:

P=1.2\times 10^{-3}kg\times 9.8m/s^2\times 1m\\\\P=1.176\times 10^{-2}J

  • Calculating the kinetic energy of coffee filter, we use the equation:

E=\frac{1}{2}mv^2

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m = mass of coffee filter = 1.2 g = 1.2\times 10^{-3}kg

v = speed of coffee filter = 0.8 m/s

Putting values in above equation, we get:

E=\frac{1}{2}\times 1.2\times 10^{-3}kg\times (0.8m/s)^2\\\\E=3.84\times 10^{-4}J

As, energy lost by coffee filter = energy gained by air molecules

So, energy lost by coffee filter = Potential energy - Kinetic energy

Energy lost by coffee filter = (1.176\times 10^{-2})-(3.84\times 10^{-4})=0.011J

Hence, the energy gained by the air molecules is 0.011 J.

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