All categories

3 years ago

Sound travels on air molecules. True False

Physics

2 answers:

natima [27]3 years ago

7 0

Answer:

I think the answer is true.

Explanation:

Ivanshal [37]3 years ago

6 0

True because sound is a type of energy made by vibrations

You might be interested in

In this diagram, medium A represents water and medium B represents air. The phenomenon in this diagram, moving from A to B, is c

Readme [11.4K]

Its C) refraction
hope its may help you

6 0

3 years ago

Read 2 more answers

The diagram shows how an image is produced by a plane mirror.

nadezda [96]

I had this question on a quiz, it was X.

3 0

3 years ago

Read 2 more answers

G How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air

sattari [20]

Answer:

686.7N

Explanation:

Given data

Mass= 70kg

We know that the buoyant force experienced by the person is equal to the weight of the person

Hence the weight is

Weight = mass* Acceleration

Weight= 70*9.81

Weight= 686.7N

Therefore the weight is 686.7N

5 0

3 years ago

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

5 0

3 years ago

In a double-slit interference experiment, the wavelength is λ = 432 nm, the slit separation is d = 0.100 mm, and the screen is d

andrew-mc [135]

In the double-slit interference experiment, the distance of the nth-maximum from the center of the screen is given by
$y= \frac{n \lambda D}{d}$
where
$\lambda$ is the wavelength
D is the distance between the screen and the slits
d is the distance between the slits

In our problem,
$\lambda=432 nm= 432 \cdot 10^{-9} m$
$D=42.0 cm=0.42 m$
$d=0.100 mm=0.1 \cdot 10^{-3} m$

By applying the previous formula, we can calculate the distance of the 4th maximum from the center of the screen:
$y_4 = \frac{(4)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=7.25 \cdot 10^{-3} m$

Similarly, the distance of the 8th- maximum is
$y_8 = \frac{(8)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=14.5 \cdot 10^{-3} m$

Therefore, the distance between the two maxima is
$\Delta y=y_8- y_4 = 14.5 \cdot 10^{-3} m- 7.25 \cdot 10^{-3} m =7.25 \cdot 10^{-3} m = 7.25 mm$

5 0

3 years ago