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Natasha2012 [34]
3 years ago
13

Is material medium required for sound waves propagation??? Give examples

Physics
1 answer:
sp2606 [1]3 years ago
7 0

Explanation:

Hey there!!!

As sound is a mechanical wave,it requires the medium to propagate. The most common waves are:

  • Air is themost common medium through which the sound travels easily.
  • Water is also a factor that transfers the sound by vibrating water molecules.
  • and solids are also medium to transfer or propagate sound waves.

<em>Hope it helps</em><em>.</em><em>.</em>

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A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho
Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

 (v-u)/t =  F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

3 0
3 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
3 years ago
A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela
larisa86 [58]

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

<span>Now we will write the equation for a Change in Potential energy ΔU,</span>

ΔU = Uf - Ui

= 0-m*g*h

<span>  ΔU = -m*g*h --Equation 2</span>

Now compare the both equation

<span>Wint = -ΔU</span>

we can rewrite the above equation

ΔU = -W.

<span>So our Answer is ΔU = -W. .</span>

<span> </span>

5 0
3 years ago
Why are there shadows?
Deffense [45]
Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.
6 0
3 years ago
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Two blocks a and b ($m_a&gt;m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the
Yuki888 [10]

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

W_a = F_a d

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

W_b = F_b d

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

F_a = F_b

and so

W_a=W_b

3 0
3 years ago
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