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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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If an object is accelerating, it is traveling the same distance for each time interval of its motion

Anika [276]

Answer:

false

Explanation:

if an object is accelerating the object will not travel the same distance every time interval

4 0

2 years ago

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what can you infer about copper and sliver based on their position relative to each other on the periodic table?

timofeeve [1]

Both are metals and are good conductors of electricity and heat.

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3 years ago

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How many valence electrons are in an atom of fluorine

Korolek [52]

Since Fluorine has 2 electrons in the s orbitals and 5 in the p orbitals of shell number 2, there is a total of 7 valence electrons.

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3 years ago

Please help me with my quiz

Oduvanchick [21]

Answer:

matter: A

data: E

variable: C

Controlled: D

Physical: B

Explanation:

8 0

2 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago