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4 years ago

Which statement is true for a sound wave entering an area of warmer air

Physics

2 answers:

Reika [66]4 years ago

8 0

That waves travel faster than the wave lenght!

kotegsom [21]4 years ago

7 0

The sound is transferred by collisions of molecules. Therefore sound waves will travel faster on warm air because collisions of molecules of air in warm air are greater.

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a gas has a volume of 8 liters at a temperature of 300 K the volume is then increased to 12 Liters what is the new temperature (

natali 33 [55]

300/8 = 37.5
37.5 x 12 = 450
New temp. = 450 K
Hope this helps!

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3 years ago

HELPP WILL MARK BRAINLIEST!!!

guapka [62]

I guess it’s B cause that maybe is the output

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3 years ago

Read 2 more answers

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

$\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

We're told the water is leaking out at a rate of $10,500\,\frac{\mathrm{cm}^3}{\rm min}$ , so we find the rate at which it's being pumped in to be

$\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}$

$\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}$

4 0

3 years ago

After the elevator accelerates, then it moves at a constant speed of 6.0m/s, calculate the tension inthe cable.

Anon25 [30]

We know that

• The mass of the elevator is 5000 kg.

Let's draw a free-body diagram.

As you can observe, there are just two forces involved, the weight of the elevator and the tension force. Let's use Newton's Second Law.

$\begin{gathered} \Sigma F_y=ma_y \\ T-W=ma_y \end{gathered}$

But, W = mg = 5000kg*9.8m/s^2 = 49,000 N, and m = 5000 kg, a = 0 (because the speed is constant).

$\begin{gathered} T-49,000N=5000\operatorname{kg}\cdot0 \\ T=49,000N \end{gathered}$ <h2>Therefore, the tension in the cable is 49,000 N.</h2>

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2 years ago

Why exactly do people use simple <br> machines like a pulley

Komok [63]

They can change the direction of force which can result in easier movement.

6 0

4 years ago