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3 years ago

These graphs show measurements of the density of air as different sound waves pass a single point.

Physics

2 answers:

inessss [21]3 years ago

7 0

Answer: The graph with the highest density of air.

Explanation: The graphs are missing, but il try to explain this problem anyways.

Soundwaves are mechanical waves, that need a medium to move (commonly, the medium is the air)

As those mechanical waves affect the air, the density of air in some areas increase and in others decrease, and when that "wave of air" impacts a receptor, like your ear, you receive the information of the sound, like pitch and intensity.

As more air impacts the receptor, more "loud" is the noise, and as you may know, density = mass/volume, so as biggest is the density of air in a point, this means that we have more mass of air at that point, which is directly related to the intensity or loudness of the sound wave at that point

Ymorist [56]3 years ago

5 0

Answer: D!

It is the option with the greatest amplitude.

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17. How long does it take a giraffe running at a speed of 33 m/s to run 200 meters !

scZoUnD [109]

Answer:

6600sec

Explanation:

T=sv

T=200m*33m/s

T=6600sec

5 0

3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

Light passes through a pair of narrow slits with a 0.67-mm separation. It is found that the fourth bright fringe makes an angle

babunello [35]

Answer:

The wavelength of the light is 555 nm.

Explanation:

according to Bragg's law..

n×λ = d×sin(θ)

n is the fringe number

λ is the wavelength of the light

d is the slit separation

θ is the angle the light makes with the normal at the fringe.

7 0

3 years ago

Grindstone: Rotational Dynamics and Kinematics You have a grindstone (a disk) that is 133.0 kg, has a 0.635 m radius, and is tur

Colt1911 [192]

Answer:

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

where N can be obtained as follows:

∑ Fc = m*ac ⇒ N - F = m*ac = m*ω²*R ⇒ N = F + m*ω²*R

then if

F = 32 N

m = 133 Kg

R = 0.635 m

ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s

we get

N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N

Finally

Ff = μ*N = 0.10*(8390.53 N) = 839.05 N

3 0

3 years ago

Amount of work done by a rotating object

Oduvanchick [21]

The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] * [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

5 0

3 years ago

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