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3 years ago

Which of the following statements is an accurate description of vibrations?

2 answers:

6 0

So we want to know what statement is an accurate description of vibrations. So humans can hear sound frequencies from 20-20000 Hz. Below 20 Hz is infra sound and above 20000 Hz is ultra sound. Humans cant hear both infra sound and ultra sound so the correct answer is A.

Daniel [21]3 years ago

6 0

Infrasonic and ultrasonic vibrations can’t be heard by humans because they’re outside of the range of frequencies that can be detected. Infrasonic vibrations have a lower frequency than can be heard by the human ear; ultrasonic vibrations have a higher frequency than can be heard by the human ear. Infrasonic vibrations can be felt if the amplitude is great enough. Ultrasonic vibrations are utilized in sonar equipment and in applications to detect flaws in steel castings and to remove grease or foreign material from machine-part surfaces.

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What is the force that keeps the Earth and other planets in their orbital paths?

Anna71 [15]

Answer:

Gravity

Explanation:

Gravity is a force that pulls the surface of the earth and keeps the planets in orbits around the sun.

7 0

3 years ago

Read 2 more answers

This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a

Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time taken for decomposition = 3 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{58}{100}\times 100=58g$

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

$k=\frac{2.303}{3}\log\frac{100}{58}$

$k=0.18days^{-1}$

a) Half-life of radon-222:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days$

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant = $0.18days^{-1}$

t = time taken for decomposition = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{15}{100}\times 100=15g$

$t=\frac{2.303}{0.18}\log\frac{100}{15}$

$t=10.54days$

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0

3 years ago

Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? S

Zepler [3.9K]

Answer:

About two kilometers away

$\rm distance=2.401\ km$

Explanation:

Given:

The time gap between the light and sound to travel to the house, $t=7\ s$

Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.

∴Distance between the lightning-strike and the house:

$\rm distance=v\times t$