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Ad libitum [116K]

2 years ago

6

A.) Review the graphical data above and create a data table that reflects the graph (10 pts.)

1 answer:

Artist 52 [7]2 years ago

7 0

No shushhhhhhhhhhhhhhhhhhh

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A trombone plays a C3 note. If the speed of sound in air is 343 m/s and the wavelength of this note is

Umnica [9.8K]

The frequency of note C3 is 131 $s^{-1}$ .

<u>Explanation:</u>

Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.

And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.

Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:

$Frequency=\frac{speed of sound}{Wavelength of note}$

Thus,

$Frequency = \frac{343 m/s}{2.62 m} = 131 s^{-1}$

So the frequency of note C3 is 131 $s^{-1}$ .

3 0

3 years ago

What is the maximum m2 value that the system can be stationary at?

Ne4ueva [31]

Answer: 37.5 kg in 3 s.f.

Explanation:

7 0

3 years ago

A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

astraxan [27]

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$

7 0

3 years ago

How does the sun create heat?

Triss [41]

Explanation:

sun was created through the hypothesis

3 0

2 years ago

Read 2 more answers

What is the force due to gravity of a 38 kg student?

alukav5142 [94]

Answer:

F_g = 372.78 N

Explanation:

Formula for force of gravity is;

F_g = mg

Where;

m is mass

g is acceleration due to gravity

We are given;

Mass = 38 kg

Acceleration due to gravity has a constant value of 9.81 m/s²

Thus;

F_g = 38 × 9.81

F_g = 372.78 N

6 0

3 years ago