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2 years ago

8

It is essential to behave appropriately on social media and be _______ when you post anything.

1 answer:

Firdavs [7]2 years ago

5 0

Answer:

I would say be Mindful.

Explanation:

There could be like a MILLION answers for this. I think that you should personally go with your gut. That would be the best option. I think it's mindful because you really do have to be mindful when you post. Like not posting too much, not posting stuff you're uncomfortable with, not posting when on vacation, etc. So, I think you should be mindful.

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A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Lunna [17]

Answer:

option A

Explanation:

given,

height of the drop of stone = 9.44 m

speed of the stone = ?

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2} m v^2$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 9.44}$

$v = \sqrt{185.024}$

v = 13.60 m/s

Hence, the correct answer is option A

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Answer:

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Explanation:

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3 years ago

your neighbour is throttling his recent bought motorbike to show off. the sound intensity measured at your window 16m away is 0.

aivan3 [116]

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3 years ago

Can someone pls help me

natali 33 [55]

Answer:

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3 years ago

Read 2 more answers

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago