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grin007 [14]
3 years ago
6

Sound waves are longitudinal waves that can travel through air. Would you expect sound waves to travel faster through a low-dens

ity gas (such as helium) or a higher-density gas such as carbon dioxide? Justify your answer based on what you have learned.
Physics
1 answer:
Serjik [45]3 years ago
6 0

Answer:

Sound waves travel faster in a low-density gas

Explanation:

First of all, let's remind that sound waves are pressure waves: they consist of oscillations of the particles in a medium, which oscillate back and forth along the direction of motion of the wave (longitudinal wave).

The speed of sound in an ideal gas is given by the equation

v=\sqrt{\gamma \frac{p}{\rho}}

where

\gamma is the adiabatic index of the gas

p is the gas pressure

\rho is the gas density

From the equation, we see that the speed of sound is inversely proportional to the square root of the density: therefore, the lower the density, the faster the sound waves.

So, sound waves will travel faster in a low-density gas.

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A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
2 years ago
Is it true or false that air pollution only occurs as a result of human activity
gregori [183]
Hi alexander it is very true air pollution is only caused by human activity

8 0
3 years ago
Read 2 more answers
Why would an insurance company want people to avoid risk, like Nick Cannon having to avoid skydiving?
Lina20 [59]
An insurance company wants people to avoid risks because if the person meets with an accsident the expense of the medicines and treatment should be covered by the insurance company.
5 0
3 years ago
Read 2 more answers
At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−
vredina [299]

Answer:

a)  Δx = 49.23 mi , b)  Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

     a = dv / dt

     ∫dv = ∫ a dt

we integrate and evaluators

      v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

       1+ 8t = u

       8 dt = du

       v - v₀ = -1280 ∫ u⁻³ du / 8

       v -v₀ = -1280 / 8 (-u⁻²/2)

       v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

      v- 80 = 80 [(1 + 8t)⁻² - 1]

      v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

     v = dx / dt

    dx = vdt

    x-x₀ = 80 ∫ (1-8t)⁻² dt

    x-x₀ = 80 ∫ u⁻² dt / 8

    x-x₀ = 80 (-1 / u)

    x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

   x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

     x = x₀

t = 0.2 h

    x-x₀ = -80 [1 / (1 +8 02) -1]

    x-x₀ = 49.23

displacement is

  Δx = x (0.2) - x (0)

   Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

     x- x₀ = -80 [1 / (1+ 8 0.4) -1]

     x-x₀ = 55 mi

    Δx = x (0.4) - x (0.2)

     Δx = 55 - 49.23

     Δx = 5.77 mi

3 0
3 years ago
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