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3 years ago

Differentiate between sound waves and seismic waves?

2 answers:

algol133 years ago

7 0

The only real difference is that common seismic waves travel through the ground and sound waves travel through the air. If you had a pipe attached to granite and you were listening to it, you might detect both.

Sloan [31]3 years ago

6 0

Seismic Waves travail under ground and Sound Waves travail trough the air.

Hope I could of help in any way.

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Plants depend on the atmosphere for the largest source of

Sonja [21]

The correct answer that would best complete the given statement above would be CARBON DIOXIDE (CO2). Plants <span>depend on the atmosphere for the largest source of carbon dioxide. Carbon dioxide is considered as an important trace gas in the Earth's atmosphere which constitutes about 0.04% of it. Hope this answer helps. </span>

6 0

3 years ago

The index of refraction of light varies from color to color. True or False?

lys-0071 [83]

The index of refraction of light varies from color to color. TRUE.

8 0

3 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

height(h)=1 m

time taken=0.64 s

$h=ut+frac{at^2}{2}$

$1=0+\frac{a(0.64)^2}{2}$

$a=4.88 m/s^2and [tex]a=\alpha r$

where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

$T=m(g-a)$

$T=1.7\times (9.8-4.88)$

T=8.364 N

and Tension will provide Torque

$T\times r=I\cdot \alpha$

$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

$I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2$

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0

3 years ago

Approximately how many atoms are there along a 9.5 cm line

uranmaximum [27]

First convert 5.5 cm to meters.

(5.5 cm / 1) x (m / 100 cm) = 0.055 m

A typical atom is about 1.0E-10 m in diameter; thus:

0.055 m / 1.0E-10 m = 5.5E8 atoms or 550,000,000 end-to-end atoms in 5.5 cm

4 0

3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$