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2 years ago

Can you give me an example of how sound passes through a solid?please help

2 answers:

5 0

When a layer of cold air close to the ground is covered by a layer of warmer air, sound waves traveling upward may be bent, or refracted, by the difference in temperature and redirected toward the ground.

alexira [117]2 years ago

3 0

Answer:

sound travels through steel about 15 times faster than through air.

Explanation:

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All of the following were barriers to minority participation in early psychology except

Masja [62]

<span>All of the following were barriers to minority participation in early psychology except school desegregation. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that you were looking for and the answer has come to your great help.</span>

4 0

2 years ago

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Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on

disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

$V = V_1 + V_2 + V_3$

here we have potential due to 50 nC placed at y = 6 m

$V_1 = \frac{kQ}{r}$

$V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_1 = 45 Volts$

Now potential due to -80 nC charge placed at x = -4

$V_2 = \frac{kQ}{r}$

$V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}$

$V_2 = -60 Volts$

Now potential due to 70 nC placed at y = -6 m

$V_3 = \frac{kQ}{r}$

$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

Now total potential at this point is given as

$V = 45 - 60 + 63 = 48 Volts$

3 0

2 years ago

Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener

qwelly [4]

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy can be given by the formula:

$f = \exp(\frac{-E_{a} }{RT} )$

$E_{a} = Activation Energy\\E_{a} = 100 kJ /mol\\E_{a} = 10^{5} J /mol\\$

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

$f_{1} = \exp(\frac{-10^{5} }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5} }{2552.398} )\\f_{1} = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}$

When Temperature = 52⁰C = 52 + 273 = 325 K

$f_{2} = \exp(\frac{-10^{5} }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5} }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}$

$\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }$

Factor = 8.77

8 0

3 years ago

The formation of a standing wave requires _____.

erica [24]

the formation of a standing wave requires interferencethe incoming and reflected waves of the same frequency

4 0

3 years ago

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A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is -1.67 m/s^2. Determine

zalisa [80]

<span>We can use a simple equation to find the time for the feather to fall to the surface. y = (1/2) g t^2 y is the height g is the acceleration due to gravity t is the time t^2 = 2y / g t = sqrt{ 2y / g } t = sqrt{ (2) (1.40 m) / (1.67 m/s^2) } t = sqrt { 1.6766 s^2 } t = 1.29 seconds It takes 1.29 seconds for the feather to fall to the surface.</span>

5 0

3 years ago