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3 years ago

Sound travels fastest in _____.

Physics

2 answers:

Irina-Kira [14]3 years ago

6 0

Sound travels fastest in solids,than liquids, and faster in liquids, than in gases.

Charra [1.4K]3 years ago

4 0

Answer:

Sound travels fastest in solids

Explanation:

Sound is a type of mechanical waves. Mechanical waves are produced by the oscillations of the particles in the medium the wave is travelling through.

The speed of a mechanical wave depends on the density of the medium: the higher the density, the higher the speed. This is because in a medium with higher density, atoms/molecules are closer together, so they collide more frequently and therefore the wave can be transmitted faster and more efficiently.

Since solids have higher density than liquids and gases, this also means that mechanical waves (so, sound as well) travel faster in solids than liquids or gases.

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What is carbon dioxide?

Bezzdna [24]

Answer:

acidic colorless gas

Explanation:

Carbon dioxide is an acidic colorless gas with a density of about 53% higher than that of dry air. Carbon dioxide molecules consist of a carbon atom covalently double bonded to two oxygen atoms. It occurs naturally in Earth's atmosphere as a trace gas.

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3 years ago

Read 2 more answers

A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevatio

egoroff_w [7]

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

$dtan13^0 = 0.231d$

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

$dtan4^0 = 0.07d$

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

7 0

3 years ago

How to use kinetic energy to calculate velocity

BartSMP [9]

As we know the formula of kinetic energy is

$KE = \frac{1}{2} mv^2$

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

$150,000 = \frac{1}{2} * 120 * v^2$

$v^2 = 2500$

$v = 50 m/s$

So speed of above object is 50 m/s

7 0

3 years ago

Can someone please help me with this? <br>

Sauron [17]

Divide 56 by 60 then x the answer by 6 then you get 5.6km which is your answer.

5 0

3 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

6 0

3 years ago