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2 years ago

Why can sounds range from barely audible to deafeningly loud?

Physics

1 answer:

slavikrds [6]2 years ago

6 0

Answer:

This is due to the frequency

Explanation:

All sounds are characterized by the frequency range they have, the frequency is the number of times a sound wave is repeated in an instant of time called period.

To better understand this concept, an image is attached in which you can see the different types of waves according to their frequency. We can see that the audible frequencies are within the range of 10^3 to 10^8 [Hz].

Outside this range sounds are imperceptible to the human ear, which is why we cannot hear sounds emitted by physical or similar phenomena. It should be noted that as the frequency of the sound wave increases, the audible capacity of the human ear decreases.

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A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m

solmaris [256]

Answer:

The strength of the magnetic field that the line produces is $2x10^{-5} Tesla$ .

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

$B = \frac{\mu_{0}I}{2\pi r}$ (1)

Where $\mu_{0}$ is the permiability constant, I is the current and r is the distance from the wire.

Notice that it is necessary to express the current, I, from kiloampere to ampere.

$I = 1.0kA \cdot \frac{1000A}{1kA}$ ⇒ $1000A$

Finally, equation 1 can be used:

$B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}$

$B = 2x10^{-5}T$

Hence, the strength of the magnetic field that the line produces is $2x10^{-5} Tesla$ .

8 0

2 years ago

Which planet has the most extreme temperature variations

vichka [17]

Answer:

I believe Mercury has the most extreme temperatures in the solar system, ranging from -280?F at night to 800 degrees F during the day for parts of the surface.

Hope that helps! :)

5 0

2 years ago

Read 2 more answers

What is the quantity of motion ??

horsena [70]

Answer:

The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly. In other words, rather than defining the quantity of motion of a given object as simply the kinematic velocity v of the object, he defined it as the product mv, where m is the mass of the object.

Explanation:

3 0

2 years ago

A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A

Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

$F = ILBsin \theta$

Here,

I = Current

L = Length

B = Magnetic Field

$\theta$ = Angle between Force and Magnetic Field

But $\theta = 90\°$

$F = ILB$

Rearranging to find the Magnetic Field,

$B = \frac{F}{IL}$

Here the force per unit length,

$B = \frac{1}{I}\frac{F}{L}$

Replacing with our values,

$B = \frac{0.130N/m}{16.5}$

$B = 0.0078T$

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0

3 years ago

Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr

slavikrds [6]

Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

$E=\frac{1}{2} *L*I^{2}$

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

$E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}$

Then:

$E'=4*E$

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

3 0

2 years ago