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Bond [772]
3 years ago
10

Which of these is a mechanical wave

Physics
1 answer:
Anika [276]3 years ago
8 0
<h2>Answer:</h2>

<h3>Sound waves</h3>

Explanation:

<h3>Mechanical Sound Waves - A sound wave moves through air by displacing air particles in a chain reaction. ... Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave.</h3>

(≚ᄌ≚)ℒℴѵℯ❤

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The critical angle for water is 49°. If a ray of light
Sonja [21]

Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49      Ni = 1.325 deg     the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg  

So refraction can occur  outside the pond and you do not have total internal refection.

 

3 0
3 years ago
A car is traveling around a horizontal circular track with radius r = 220 m at a constant speed v = 16 m/s as shown. The angle θ
iogann1982 [59]
Velocity = distance / time = ( 2 * pi * r ) / t = 20.583 m/s 

<span>x component = sine ( 32 ° ) * 20.583 = 10.91 m/s

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5 0
2 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
2 years ago
Find the frequency of a wave of wavelength 2.5m and speed 400 m/s
Sedaia [141]

Answer:

The frequency of wave is 160Hz.

Explanation:

Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.

So first thing, you have to make frequency the subject by dividing wavelength on both sides :

v = f \times λ \:

v \div  λ =  f \times λ \div λ

f =  \frac{v}{λ}

Next you have to substitute the value of v and f into the formula :

Let λ = 2.5m,

Let v = 400m/s,

f =  \frac{400}{2.5}

f = 160Hz

7 0
2 years ago
A solenoid field is:
Bumek [7]

increased with an increased current flow

7 0
3 years ago
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