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3 years ago

Sounds travels in a ____ wave A. Transverse B. Compressional C. Surface D. Inverted

Physics

1 answer:

MissTica3 years ago

7 0

Answer: A. Transverse

Explanation:

Sound is a mechanical transverse wave, it travels faster in solids than in liquids or gases. This is because the speed of the mechanical waves is determined by a relationship between the elastic properties of the medium in which they are propagated and the mass per unit volume of the medium.

In addition, the speed of sound varies with changes in the temperature of the medium. This is because an increase in temperature means that the frequency of interactions between the particles that transport the vibration increases, hence this increase in activity increases the speed.

Hence:

<h3>Sounds travels in a <u>transverse</u> wave</h3>

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3 resistors of 3,4 and 12 ohm are there. How would you connect so as to get a resistance more than 5 oh, but less than 7 ohm sho

DochEvi [55]

Answer:

<em>Connecting the 4-ohm and 12-ohm in parallel and followed by the 3-ohm resistor in series. A scheme of the configuration is attached below. Please see the file attached below to know the diagram.</em>

Explanation:

There is the following solution that satisfies all requirements indicated on statement:

<em>Connecting the 4-ohm and 12-ohm in parallel and followed by the 3-ohm resistor in series. A scheme of the configuration is attached below.</em>

$R = \frac{(12\,\Omega)\cdot (4\,\Omega)}{12\,\Omega+4\,\Omega}+3\,\Omega$

$R = 6\,\Omega$

Which observes all design requirements.

4 0

3 years ago

Can someone please help me because this is a hard problem to solve.

ivolga24 [154]

Answer:

25.0 m / 10 m/s = 2.5 s

50.0 / 9.5 = 5.26 s

25.0 / 11.1 = 2.25 s

T (Wood) = 2.5 + 5.26 + 2.25 = 10.0 s

Mrs Wood runs 10 s vs 10.4 for Mr Overstreet

7 0

3 years ago

a city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in

ivolga24 [154]

We have that for the Question "A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?" it can be said that

OC=2

From the question we are told

A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?

Generally the equation for the Opportunity cost is mathematically given as

$OC=\frac{y_1-y_2}{x_1-x_2}\\\\OC=\frac{2-0}{1-0}\\\\OC=2$

Therefore

From the graph of the question we can ascertain that Opportunity cost

OC=2

For more information on this visit

brainly.com/question/18670421

Graph is attached below

6 0

2 years ago

Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10

daser333 [38]

3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

= 64998 rad/min

Formula used,

$v=rw$ where,

$v$ is the Linear velocity

$r$ is the radius

$w$ is the angular velocity

$v$ = 6 cm × 64998rad/min

= 3,89,988 cm/min

Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.

Learn more about Angular speed here brainly.com/question/540174

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8 0

2 years ago

Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside

jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

$\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}$

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

$\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}$

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0

1 year ago