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3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Physics

1 answer:

PolarNik [594]3 years ago

6 0

OPTION C is the correct answer.

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The number of wavelengths that pass a given point/second is called

myrzilka [38]

Answer:

I believe the answer is frequency

Hope this helps!! (let me know if it is right)

Explanation:

6 0

3 years ago

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outst

Contact [7]

Answer:

$v_{i}=10.10 m/s$

Explanation:

The equation of the position is:

$y=y_{i}+v_{i}t-0.5gt^{2}$

Where:

v(i) is the initial velocity

The initial position y(i) will be zero and the final position y = 3.60 m.

So, we just need to solve this equation for v(i).

$v_{i}=\frac{y+0.5gt^{2}}{t}$

$v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}$

$v_{i}=10.10 m/s$

Therefore, the initial velocity is 10.10 m/s upwards.

I hope it helps you!

5 0

3 years ago

A typical adult human lung contains about 330 million tiny cavities called alveoli. Estimate the average diameter of a single al

Amanda [17]

Answer:

The average diameter of a single alveolus is 0.0222 cm.

Explanation:

Volume of the lung ,V= 1.9 L

$1 L = 1000 cm^3$

$1.9 L=1.9\times 1000 cm^3=1900 cm^3$

Number of alveoli in a human lung = $330\times 10^6$

Volume of single alveoli =v

$v\times 330\times 10^6=V$

$v=\frac{1900 cm^3}{330\times 10^6}$

$v=5.7575\times 10^{-6} cm^3$

The alveoli are spherical.

Radius of an alveolus = r

Volume of the sphere = $\frac{4}{3}\pi r^3$

$v=\frac{4}{3}\pi r^3$

$5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3$

$r=0.0111 cm$

Diameter of the alveolus =d

d = 2r = 2 × 0.0111 cm = 0.0222 cm

The average diameter of a single alveolus is 0.0222 cm.

7 0

3 years ago

Organ pipe a, with both ends open, has a fundamental frequency of 340 hz. the third harmonic of organ pipe b, with one end open,

Sphinxa [80]

The fundamental frequency of the open pipe A
Length 'L₀' is, f₀ = U/2L₀ = 340 H₂
the speed of sound in air V= 343m/s
∴343/2L₀ = 340 → length L₀ = 343/2 ×340 = 0.5044 = 50.44
The third harmonic of closed pipe 'B' is
F3₀ = 3V/4LC
The second harmonic of open pipe 'A' is
f2₀ = 2V/2L₀
∴ 3V/4LC = 2V/2L →L₀ = 3L₀/4
⇒ The length of closed pipe B is
Lc = 37.83cm

7 0

3 years ago

A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpt

NikAS [45]

We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>

8 0

3 years ago

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