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3 years ago

When large amplitude sound vibrates the ear significantly, a _______ results.

2 answers:

7 0

The answer is D. The loudness of a sound, or volume of a sound wave, can be dictated by the abundance of the sound wave. Both pitch and volume can influence how "boisterous" a sound is and how it influences human ears particularly.

kondaur [170]3 years ago

4 0

D loud sound

Hope this helped:)

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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

Nat2105 [25]

Answer:

Explanation:

Given

diameter $d=20 \mu m$

density $\rho =1300 kg/m^3$

frequency $\nu =150 Hz$

Length of silk strand $L=14 cm$

Velocity in the string is as follows

$\nu =\sqrt{\frac{T}{\mu }}$

The expression for Fundamental Frequency

$f=\frac{\nu }{2l}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}$

$f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}$

Squaring

$f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}$

$T=4\rho \cdot A\cdot l^2\cdot f^2$

$T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2$

$T=7.2\times 10^{-4} N$

8 0

4 years ago

A Truck with a mass of 1500 kg is decelerated At a rate of 5m/s2. how much force did this require

Marianna [84]

(1500 kg)*(5 m/s^2) = 7500 N

6 0

3 years ago

Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long

NNADVOKAT [17]

Answer:

a) 578.0 cm²

b) 25.18 km

Explanation:

We're given the density and mass, so first calculate the volume.

D = M / V

V = M / D

V = (6.740 g) / (19.32 g/cm³)

V = 0.3489 cm³

a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.

V = Ah

A = V / h

A = (0.3489 cm³) / (6.036×10⁻⁴ cm)

A = 578.0 cm²

b) The volume of a cylinder is pi times the square of the radius times the length.

V = πr²h

h = V / (πr²)

h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)

h = 2.518×10⁶ cm

h = 25.18 km

3 0

3 years ago

A student pulls a cart across the floor with a force of 45 N. If the cart travels 10 m, how much work does the cart do on the st

bixtya [17]

I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.

6 0

3 years ago

A uniform magnetic field of magnitude 0.23 T is directed perpendicular to the plane of a rectangular loop having dimensions 7.8

Arlecino [84]

Answer:

0.0025116weber/m²

Explanation:

Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).

Mathematically;

B = ¶/A

¶ = BA

Given B = 0.23Tesla which is the magnitude of the magnetic field

Dimension of the rectangular loop = 7.8 cm by 14 cm

Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm

= 109.2cm²

Converting this value to m²

Area of the loop = 109.2 × 10^-4

Area of the loop = 0.01092m²

Magneto flux = 0.23×0.01092

Magnetic flux = 0.0025116weber/m²

3 0

3 years ago

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