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3 years ago

Sound waves move the fastest through which medium

Physics

1 answer:

chubhunter [2.5K]3 years ago

3 0

Answer:

water at high temperature

Explanation:

When heated, molecules move faster which results in the sound waves being transmitted faster.

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Describe the factors that cause static friction between two surfaces to increase.

maw [93]

Answer : The static friction depends on two factors

1. Roughness of the surface

2. Force

Explanation : Friction occur when surface is not smooth.

The formula of friction is

$F= \mu mg$

Static friction depends on the roughness of the surface and force which is trying to push to object along the surface.

Static friction is caused by the attraction between two surfaces that are in contact. when the surface will rough and the object will heavier then the force will be larger.

4 0

3 years ago

The force exerted by the earth on a body is called ______of a body.

AVprozaik [17]

Answer:

The force exerted by the earth on a body is called gravitational of a body.

i think this is the answer

4 0

3 years ago

Read 2 more answers

a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec

Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound $s$ is given by a relation between the distance $d$ and the time $t$ :

$s=\frac{d}{t}$ (1)

Where:

$s=330 m/s$ is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

$d=412.5 m-x$ since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance $x$ towards the cliff

$t=\frac{2.2 s}{2}=1.1 s$ Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate $d$ and then find $x$ :

$d=st$ (2)

$412.5 m-x=(330 m/s)(1.1 s)$ (3)

Finding $x$ :

$x=49.5 m$ This is the distance at which the hunter is from the cliff.

3 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

When a guitar string plays the note "a," the string vibrates at 440 hz ?

Rama09 [41]

Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.

8 0

2 years ago