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3 years ago

13

Sound is a mechanical wave; therefore, it __________.

2 answers:

avanturin [10]3 years ago

8 0

A mechanical wave such as sound needs a medium to travel through. This is why they say that in space, no one can hear you scream (because space is a vacuum and no medium exists). On the contrary, light can exist in wave or particle form. This is why we have the ability to see beyond the realms of the Earth.

miss Akunina [59]3 years ago

6 0

Answer:

The answer is A. Must travel trought a medium.

Explanation:

The transmission of sound depends on the existence of a material medium (solid, liquid, air particles) where the vibration of the molecules are perceived as a sound wave, which are converted into mechanical waves in the human ear, to be later perceived by the brain.

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The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes it 0.90 s to reach the floor. Wha

Anna007 [38]

Answer:

The mass of the another block is 60 kg.

Explanation:

Given that,

Mass of block M= 100 kg

Height = 1.0 m

Time = 0.90 s

Let the mass of the other block is m.

We need to calculate the acceleration of each block

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$1.0=0+\dfrac{1}{2}\times a\times(0.90)^2$

$a=\dfrac{2\times1.0}{(0.90)^2}$

$a=2.46\ m/s^2$

We need to calculate the mass of the other block

Using newton's second law

The net force of the block M

$Ma=Mg-T$

$T=Mg-Ma$ ....(I)

The net force of the block m

$ma=T-mg$

Put the value of T from equation (I)

$ma=Mg-Ma-mg$

$m(a+g)=M(g-a)$

$m=\dfrac{M(g-a)}{(a+g)}$

Put the value into the formula

$m=\dfrac{100(9.8-2.46)}{2.46+9.8}$

$m=59.8\ \approx60\ kg$

Hence, The mass of the another block is 60 kg.

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3 years ago

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The correct answer is 4

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vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

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$v=\omega A$

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A=0.075 m

A= 0.75 cm

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$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

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Gala2k [10]

Answer:

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Explanation:

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3 years ago

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