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2 years ago

How fast do sound waves travel in air? How does this compare to the speed of light waves?

1 answer:

3 0

Soundwaves are longitudinal waves whereas light (electromagnetic waves) are transverse waves.
Soundwaves oscillate parallel to the direction of propagation.
Light waves oscillate at right angles to the direction of propagation.

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A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont

PIT_PIT [208]

Answer:

$v' = 2.83 m/s$

Explanation:

Velocity of wave in stretched string is given by the formula

$v = \sqrt{\frac{T}{\mu}}$

here we know that

T = 4 N

also we know that linear mass density is given as

$\mu = 1 kg/m$

so we have

$v = \sqrt{\frac{4}{1}} = 2 m/s$

now the tension in the string is double

so the velocity is given as

$v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s$

$v' = 2.83 m/s$

4 0

3 years ago

Why are there shadows?

Deffense [45]

Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.

6 0

3 years ago

Read 2 more answers

What conditions can create the largest waves in the ocean. this is for science. i need done asap.

Yuki888 [10]

The answer is strong winds, i hoped this helped.

→if this helped please mark brainliest i need to level up←

7 0

3 years ago

A stationary 90 kg baseball pitcher throws a 0.15 kg baseball forward

Marina CMI [18]

Answer: momentum = 6kgm/s

Explanation:

given that the baseball pitcher is at stationary position, his velocity will be equal to zero. If velocity is zero, his linear momentum will therefore equal to zero.

Linear momentum is the product of mass and velocity. Given that the baseball has

Mass M = 0.15 kg

Velocity V = 40 m/s

Momentum = MV

Momentum = 0.15 × 40 = 6 kgm/s

4 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