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3 years ago

When two waves in the same medium hit each other, the resulting displacement of the medium is

Physics

1 answer:

Irina-Kira [14]3 years ago

7 0

Answer:

1. Either larger or smaller than the displacement of either wave acting alone, depending on the signs of the displacements of the two waves.

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State one of Kepler's laws of gravitation

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All planets move around the sun

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3 years ago

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Help!!! Line B touches the circle at a single point. Line A extends through the center of the circle.

Ulleksa [173]

Answer:

If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.

The size of the circle determines.
The diameter should not be fixed either.

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3 years ago

The equatorial diameter of venus is 7,523 miles. if a mile equals 1.609 km, what is venus's diameter in kilometers?

polet [3.4K]

The diameter of venus in km is 12104.507Km.

<h3 /><h3>What is Unit conversion?</h3>

By definition, unit conversion refers to the division or multiplication operation used to convert measurements of the same quantity between various units. The act of converting something from one form to another in mathematics, such as from inches to millimetres or from litres to gallons, is known as conversion.

the diameter of venus = 7,523 miles

1 mile = 1.609 km

so,

diameter of venus = 7523 × 1.609 Km

= 12104.507Km

to learn more about unit conversion go to - brainly.com/question/13016491

#SPJ4

6 0

1 year ago

I need help fast please help

Irina18 [472]

A and C Im pretty sure :)

5 0

3 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

3 years ago