Scientific endeavor is sometimes accompanied by negative tradeoffs.<br> true or false???????????? :0

The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

so

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

3 0

3 years ago

Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line

Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

$y=\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

$\lambda=632.8 nm = 632.8\cdot 10^{-9} m$ is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

$n=520 lines/mm$ is the number of lines per mm, so the spacing between two lines is

$d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m$

Therefore, substituting m = 1, we find:

$y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m$

So, on the distant screen, there is 1 bright fringe every 33 cm.

6 0

3 years ago

A uniform electric field contains a number of particles. All are experiencing forces in the same direction as the electric field

myrzilka [38]

Answer: D

All the particles must be uncharged

Explanation:

If all the particles are positively charged, then there will be force of repulsion between them which will give different directions away from each other. The same is applicable if they are all negatively charged.

If the particles are positively and negatively charged, their will be force of attraction between them which will give different directions towards each other.

For all to be experiencing forces in the same direction, We can conclude that

All the particles must be uncharged.

8 0

3 years ago

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$