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shepuryov [24]
3 years ago
10

Which of these factors will increase the speed of a sound wave in air?

Physics
1 answer:
Artemon [7]3 years ago
3 0
<h3>Answer;</h3>

-Temperature

<h3><u>Explanation;</u></h3>
  • Sound is a type of mechanical wave, which means it requires a material medium for transmission. It results from the vibration of particles.
  • The speed of sound in mediums varies depending on the property of the medium and a number of other factors which includes; temperature, pressure,  and humidity.
  • Temperature increases the speed of sound wave as particles at higher temperatures tend to possess more energy and thus they will vibrate faster and thus the sound wave will travel faster.
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A 4 kW vacuum cleaner is powered by an electric motor whose efficiency is 90%. (Note that the electric motor delivers 4 W of net
RoseWind [281]

Answer:3.6\ kW

Explanation:

Given

Power Supplied [tex]P_{input}=4\ kW[/tex]

Efficiency of the motor \neta =90\%

and \neta =\dfrac{\text{Power output}}{\text{Power input}}

\Rightarrow 0.9=\dfrac{P_{out}}{4}

\Rightarrow P=0.9\times 4

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3 0
3 years ago
Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

g=9.8 m/s^2 is the acceleration due to gravity

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
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Answer:

0.749 m

Explanation:

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Answer:

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Explanation:

I am 90% sure this is correct but if I'm not please tell me

3 0
3 years ago
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