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lorasvet [3.4K]
3 years ago
14

Human fingers are very sensitive, detecting vibrations with amplitudes as low as 2.0x10^-5m. Consider a sound wave with a wavele

ngth exactly 1000 times greater than the lowest amplitude detectable by fingers. What is this waves frequency​
Physics
1 answer:
Kobotan [32]3 years ago
3 0

Answer:

18,850 Hz

Explanation:

We need to figure out the wavelength of the sound wave.

Thus,

Wavelength = 1000 * Lowest Amplitude Wave

Wavelength = 1000 * 2.0 * 10^(-5)

Wavelength = 0.02

Or,

\lambda=0.02

Now, we need the frequency of this wave. It goes by the formula:

f=\frac{v}{\lambda}

Where

f is the frequency in Hz

v is the speed of sound in air (to be 377 m/s)

\lambda is the wavelength (we found to be 0.02)

Substituting, we find the frequency:

f=\frac{v}{\lambda}\\f=\frac{377}{0.02}\\f=18850

The wave has frequency of 18,850 Hz

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What wave is a wave in which the vibration of the medium is parallel to the direction the wave travels.?
galina1969 [7]
The answer to this question would be a transverse wave, because the vibration travels parallel to the direction that the wave is traveling.
7 0
2 years ago
Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica
tiny-mole [99]

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

        Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

        Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         ½ k x² + m g x ’= ½ m v²

         v² = \frac{k}{m}  x² + g x

let's calculate

         v² = \frac{8700}{55}  1.25² + 9.8 1.25

         v² = 247.159 + 12.25 = 259.409

         v = 16.11 m / s

8 0
3 years ago
Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2
Mrrafil [7]

Answer:

The girl exerts more pressure.

Explanation:

Pressure can be defined as the force exerted normally or perpendicularly per unit area.

i.e P = F/A

<u>Girls</u>

Area of the heel = 1cm² = 10^(-4) m²

Force = mg = 50 × 10 = 500N

Pressure =

\frac{500}{10 ^{ - 4} }

= 5 \times  {10}^{6}

<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u>\frac{40000}{2.5 \times  {10}^{ - 2} }</u>

<u>= 1.6 \times  {10}^{6}</u>

5 0
3 years ago
At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is
n200080 [17]

a) 8.93\cdot 10^{-30} N, downward

b) 1.76\cdot 10^{-17} N, upward

c) 1.20\cdot 10^{-17} N, downward

Explanation:

a)

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m=9.11\cdot 10^{-31} kg is the mass of the electron

g=9.8 m/s^2 is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N

And the direction is downward, towards the Earth's centre.

b)

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

F=qE

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

q=1.6\cdot 10^{-19} C is the charge of the electron

E=110 N/C is the strength of the electric field

Substituting,

F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N

And since the electron has negative charge, the  direction of the force is opposite to that of the electric field, so upward.

c)

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

F=qvB

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

q=1.6\cdot 10^{-19} C is the charge of the electron

v=1.50\cdot 10^6 m/s is the velocity

B=50.0\mu T = 50.0 \cdot 10^{-6} T is the strength of the magnetic field

Substituting,

F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

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2 years ago
Describe the heliocentric model of the universe
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