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3 years ago

11

SOUND Moves through a (blank) or (blank) material when the atoms within the material push against each other , transferring the

energy from one particle to the next

1 answer:

harina [27]3 years ago

7 0

It is probably transversal.
The particles vibrate perpendicular to the direction of an energy transmission.
I hope this helps.

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What is the definition of a volt in science?

OverLord2011 [107]

<span> One </span>volt<span> is </span>defined<span> as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points.</span>

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3 years ago

The periodic wave in the diagram below has a frequency of 80. hertz.

ASHA 777 [7]

Answer:

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3 years ago

The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4

NNADVOKAT [17]

The position of the mass is given by (in cm):
$x(t)=2 \cos (10 t)$
The velocity is the derivative of the position:
$v(t) = \frac{dx(t)}{dt} =-10\cdot 2 \sin (10t)=-20 \sin (10t)$
Substituting t=0.40 s, we can find the velocity at this time:
$v(0.40 s)= -20 \sin (10 \cdot 0.4)=15 cm/s=15 \cdot 10^{-2}m/s$

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3 years ago

Read 2 more answers

A car is moving down the street at 53km/h. A child suddenly runs into the street. If it takes the driver 0.75 sec to react and a

BARSIC [14]

Answer:

11.04m

Explanation:

Convert kilometers to meters: 53km/h x 1000 = 53,000m/h

Convert hours to seconds: 53m/h / 3600 =14.72m/s

14.72 (.75) = 11.04

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3 years ago

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,

beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

$\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2$

∴ Average intensity is 0.477 W/m²

b) Rms value

$\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C$

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

$E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C$

∴ Peak value of the electric field is 18.96 N/C

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3 years ago