Please need help on this I know it’s not D an atom.

In a compression wave, particles in the medium move

MissTica

<span>The particles through which compressional waves travel move in the same direction as the wave. This may be observed by fixing one end of a large spring and then compressing and extending the other end. The wave travels from one end to the other and the spring's parts move in the same direction.</span>

7 0

3 years ago

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When the heat source is removed from a fluid, convection currents in the fluid will eventually ____________then stop.

Blizzard [7]

When the heat source is removed from a fluid, convection currents in the fluid will eventually distribute heat uniformly throughout the fluid. When all of the fluid is at the same temperature, convection currents will stop.

5 0

4 years ago

The scale model of an airplane is 1:130, where 1 inch on the model is equal to 130 inches on the actual airplane. Which statemen

Delicious77 [7]

Answer:

The \: parts \: of \: the \: model \\ airplane \: are \: in \: the \\ same \: proportions \: as \: the \\ actual \: airplane.

3 0

3 years ago

What colors are in a 24 pack of crayola crayons?

Xelga [282]

Blue puprle white gold organs yellow red gray brown green pink black tan light blue

3 0

4 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(c) v = 0.31m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

4 years ago