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Natasha_Volkova [10]
3 years ago
9

A piano emits frequencies that range from a low of about 28 hertz to a high of about 4200 hz.find the range of wavelength in air

attained by this instrument when the speed of sound in air in 342m /s
Physics
1 answer:
AVprozaik [17]3 years ago
3 0
The solution for this problem is:
A velocity of wave is given as V = λ⋅f
V - velocity of wave
f - frequency of wave
λ -  wave length

So getting the wave length is:
λ = V/f
λminimum = V / fmaximum
λminimum = 342 / 4200
λminimum = 0.081 m

λmaximum = V/ fminimum
λmaximum = 342 / 28
λmaximum = 12.214 m
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A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
3 years ago
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At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

h=\dfrac{3v^2}{8g}

So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

6 0
3 years ago
What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two
lesya [120]

Answer:

Power  = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

\beta (dB)= 10 \  log_{10 } (\dfrac{I}{I_o})

where;

the threshold of hearing   I_o = 10^{-12} (W/m^2)

\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})

10^{10.2} =  \dfrac{I}{10^{-12}}

I = 10^{10.2} \times 10^{-12}

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power  = 124.50 W

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Pressure is the amount of force exerted on an object and force is strength or energy of an action
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2 years ago
This model shows an example of a fog bank formation. This can happen in the Great Lakes area as warm summer air moves across the
Vadim26 [7]

Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.

<h3>How Fog banks formed?</h3>

Fog banks form at sea where cool air moves quickly over the surface of the ocean that is warm. The cool incoming air lowers the temperature of the air just above the water surface and water vapor condenses into fog.

So we can conclude that Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.

Learn more about fog here: brainly.com/question/18943608

#SPJ1

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2 years ago
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