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olganol [36]
3 years ago
15

The trough of the sine curve used to represent a sound wave corresponds to

Physics
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

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Which explanations provide support for continental drift theory? Check all that apply.
denis23 [38]
<h3><u>Full Question:</u></h3>

Which explanations provide support for continental drift theory? Check all that apply.

Dinosaurs lived on many continents.

Coal fields match up across continents.

Tropical plant fossils were found in Arctic areas.

Similar rock types are found across continents.

Evidence of glaciers can be found in South Africa.

Coal fields match up across continents.

Similar rock types are found across continents.

Tropical plant fossils were found in Arctic areas.

Evidence of glaciers can be found in South Africa.

<h3><u>Explanation:</u></h3>

The way the continents on the earth shifts its position is explained by the Continental drift theory. Alfred Wegener was the one who fist proposed this theory in 1912. This theory also explains how similar rocks are formed on the different continents and also  the reason why some of the animal and plant fossils looks similar.

The evidences that supports the continental drift theory can be Coal fields match up across continents. Fossils of Glossopteris supports the  theory of continental drift with coal fields  and coastlines . The land features, climate changes are some of the evidences that also supports the theory of continental drift.The evidences of the glaciers that can be found in South Africa also supports the explanation of the continental drift.

8 0
3 years ago
A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera
yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

\Delta V_{max} = 240 V

frequency, f = 50Hz

I_{max} = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, \omega =  2\pi f = 2\pi \times50 = 100\pi

a). Inductive reactance,  X_{L} is given by:

\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega

X_{L} = 47.12\Omega 

b). The capacitive reactance,  X_{C} is given by:

\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega

X_{C} = 0.636\Omega

c). Impedance, Z = \frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega

Z = 2400\Omega

d). Resistance, R is given by:

Z = \sqrt {R^{2} + (X_{L} - X_{C})}

2400^{2} = R^{2} + (47.12 - 0.636)^{2}

R = \sqrt {5757839.238}

R = 2399.5\Omega

e). Phase angle between current and the generator voltage is given by:

tan\phi = \frac{X_{L} - X_{C}}{R}

\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})

\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}

\phi = 1.11^{\circ}

5 0
4 years ago
Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is
goblinko [34]

Answer:

d \frac{x}{l} = m×  λ⇒ d = λ ×m×l / x

= 630×10^{-9} m × 3×3m/ 45×10^{-3} m

= 1.26×10^{-4}m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0
3 years ago
Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
3 years ago
Any help????????????
pashok25 [27]
The answer should be D
3 0
3 years ago
Read 2 more answers
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