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1 year ago

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When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

otential energy? It decreased by a factor of 3. It decreased by a factor of 2. It increased by a factor of 2. It increased by a factor of 3.

1 answer:

Gnesinka [82]1 year ago

6 0

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

$U=mgh$

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

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a) 4.5 ms

The period of a wave is given by:

$T=\frac{1}{f}$

where f is the frequency.

For the note in this problem, f = 220 Hz, so the period of the wave is

$T=\frac{1}{f}=\frac{1}{220 Hz}=4.5\cdot 10^{-3} s = 4.5 ms$

b) 1381.6 rad/s

The angular frequency is given by:

$\omega=2 \pi f$

where f is the frequency.

In this problem, f = 220 Hz, so the angular frequency is

$\omega=2 \pi (220 Hz)=1381.6 rad/s$

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The frequency of the "high A" is four times the frequency of the piano string, so

$f=4 \cdot 220 Hz=880 Hz$

And so, its period is

$T=\frac{1}{f}=\frac{1}{880 Hz}=1.1\cdot 10^{-3} s=1.1 ms$

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$\omega=2 \pi f$

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For this note, f = 880 Hz, so the angular frequency is

$\omega=2 \pi (880 Hz)=5526.4 rad/s$

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1 year ago

Calculate the critical angle between glass (n = 1.90) and ice (n = 1.31)? 43° 52° 60° 75°

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Answer: $43 \°$

Explanation:

According to <u>Snell’s Law</u>:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$

Where:

$n_{1}=1.90$ is the first medium index of refraction (glass)

$n_{2}=1.31$ is the second medium index of refraction (ice)

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

Now, the critical angle $\theta_{c}$ is the angle from which there is no refraction and all the incident light is reflected to the same medium from which it proceeds, that is, the total internal reflection occurs. This is only possible when the index of refraction of the medium where the light strikes is higher than the index of refraction of the other medium, then the second angle (the exit angle) will reach the $90\º$ , for this critical incident angle $\theta_{c}$ .

Since $n_{1}>n_{1}$ , $\theta_{1}=[tex]\theta_{c}$ [/tex] and $\theta_{2}=90\º$ , hence:

$n_{1}sin(\theta_{c})=n_{2}sin(90\º)$

$(1.90)sin(\theta_{c})=(1.31)sin(90\º)$

Isolating $\theta_{c}$ :

$\theta_{c}=sin^{-1} (\frac{1.31}{1.90})$

Finally:

$\theta_{c}=43.5\º$

The option that is close to this value is $43\º$

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Answer:

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