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3 years ago

Is a measure of how closely packed together the particles of matter are in a specific volume

Physics

1 answer:

Mnenie [13.5K]3 years ago

5 0

Answer:

Density is an important physical property of matter. It reflects how closely packed the particles of matter are. When particles are packed together more tightly, matter has greater density.

Explanation:

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Which waves require a material medium for transmission? a.) light waves b.) radio waves c.) sound waves d.) microwaves

shepuryov [24]

Answer:

mechanical waves require a medium

5 0

3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

In a building with 10.000 cubic feet where the air changes every two hours, what the rate of air change? A. 167.7 cfm B. 83.3 cf

Naya [18.7K]

Answer:

Flow rate of air is given as 83.33 cubic feet per minute

Explanation:

As we know that total volume of the air flow is given as

$V = 10,000 cubic feet$

also we know that total time is

$t = 2 hours = 120 min$

now we have flow rate given as

$Q = \frac{V}{t}$

$Q = \frac{10000}{120}$

$Q = 83.3 cf/m$

3 0

3 years ago

Without Newton’s third law of motion, what would an object sitting on a table do?

Angelina_Jolie [31]

Newton's third law: If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

8 0

3 years ago

Read 2 more answers

Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y

Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density $\rho = 1000 kg/m^3$ , and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool