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3 years ago

The speed of sound in humid air than in dry air. Why

Physics

1 answer:

Elden [556K]3 years ago

3 0

Answer:

When moisture is removed from air, its density increases. The speed of sound in a humid air is inversely proportional to the square root of its density. Therefore, the speed of sound in moist air is humidity increases, the velocity of sound increase and vice-versa.

Hope it helps :)

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You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie

Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance. i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by = $\frac{1}{0.06^2}$

= $3.6 \times 10^{-3}$ times

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3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is <u>6.4m</u>

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3 years ago

Calculate how much work is required to launch a spacecraft of mass mm from the surface of the earth (mass mEmE, radius RERE) and

SIZIF [17.4K]

Answer:

Work done = (1/2)[(Gmm_e)/(R_e)]

Explanation:

I've attached the explanations below.

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3 years ago

Read 2 more answers

A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

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2 years ago

The energy transfer diagram represents the energy of a person diving into a pool.

GREYUIT [131]

The answer is A) 1000 J
According to the law of conservation of energy, the total energy before an action is always equal to the total energy after the action.
So that is, the total energy is 8000J found as potential energy, 7000J has transformed into kinetic energy, then the thermal energy should be the remaining 1000J.

Hope this helps.

6 0

3 years ago

The speed of sound in humid air than in dry air. Why ​

The speed of sound in humid air than in dry air. Why