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Kazeer [188]
3 years ago
14

Hans observed properties of four different waves and recorded observations about each one in his chart.

Physics
2 answers:
bogdanovich [222]3 years ago
8 0
The answer to your question is (Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.)


HOPE THIS HELPS YOU CAUSE IM DOING TEST AND I HOPE IM RIGHT AS WELL

PLZ GIVE ME THE BRAINLIEST ANSWER HAVE A GOOD DAY




weeeeeb [17]3 years ago
6 0

The correct answer is Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.

Wave X is bouncing off the surface thats why it is being reflected. As the waves pass from one medium to another they change speed i.e. when passing through water into air. This causes waves to bend and change color hence waves X and Z. The scattering of waves through a small opening is called diffraction so wave W

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Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will
djyliett [7]

Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

P\cdot V = n \cdot R_{u}\cdot T (1)

Where:

P - Pressure.

V - Volume.

n - Molar quantity, in moles.

R_{u} - Ideal gas constant.

T - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

P_{1}\cdot V_{1} = P_{2}\cdot V_{2} (2)

If we know that \frac{V_{2}}{V_{1}} = \frac{1}{3}, then the resulting pressure of the system is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 3\cdot P_{1}

The resulting pressure is 3 times the initial pressure.

4 0
3 years ago
Which of the following explanations is correct about the wave that travels through the medium?
SVETLANKA909090 [29]
The correct answer is letter A
5 0
3 years ago
To make a given sound seem twice as loud, how should a musician change the intensity of the sound?
Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0
3 years ago
Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a
DaniilM [7]

Answer:

a) k = 1343.6\,\frac{N}{m}, b) l = 0.501\,m\,(50.1\,cm)

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

F \propto \Delta l

In this case, the force is equal to the weight of the object:

F = m\cdot g

F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 80.614\,N

The spring constant is:

k = \frac{F}{\Delta l}

k = \frac{80.614\,N}{0.408\,m-0.348\,m}

k = 1343.6\,\frac{N}{m}

b) The length of the spring is:

F = k\cdot (l-l_{o})

l = l_{o} + \frac{F}{k}

l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }

l = 0.501\,m\,(50.1\,cm)

6 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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