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konstantin123 [22]
3 years ago
11

Do compressional waves need a medium? If so which one would be the fastest ans slowest medium.

Physics
1 answer:
Nastasia [14]3 years ago
5 0
Yes, compression waves need a medium. without a medium there is nothing to compress, hence, no wave. A fast medium like a gas (air) is easy to compress and allows waves to move through it easily. a slow medium, like a liquid, is still pretty fast, but not as fast as air.
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 <span>2IBr ⇄ I2 + Br2
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Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.<span> You can calculate the equilibrium concentration if you know the equilibrium constant Kc (Kc=I^2*Br^2/IBR^2) and the initial concentration for the reaction. The initial concentration is obtained from ICE Table.</span>
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ANSWER ASAP Which factor determines the amount of gas that can dissolve in ocean water? (A) the motion of the air above the ocea
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The amount of each gas that can dissolve in the ocean depends on the solubility and saturation of the gas in water. Solubility refers to the amount of a dissolved gas that the water can hold under a particular set of conditions, which are usually defined as 0o C and 1 atmosphere of pressure.

Explanation:

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The sizes of most of the cells our bodies are about A.1micrometer
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1Mm

Explanation:

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2 years ago
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Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do
Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

\theta=90-40=50^o

From the triangle OAB

cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}

1340.57=35^2+25^2-x^2

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0
3 years ago
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