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3 years ago

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What is mechanical advantage?

1 answer:

Dmitry_Shevchenko [17]3 years ago

7 0

Answer:

0A.Output force divided by input force.

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Determine whether the variable is qualitative or quantitative. Favorite sport. Is the Variable Qualitative or Quantitative?

Over [174]

The variable is qualitative,
the quantitative variables are those that can be specified by a numeric value.

4 0

3 years ago

The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light.

Vladimir79 [104]

If you do not have to use relative physics but classic physics, this is how you solve it:

Speed of light = c = 3 * 10^5 km/s

Speed of your foe respect to you: 0.259c

Speed of the torpedo respect to you: 0.349c

Speed of the torpedo respect your foe: 0.349c - 0.259c = 0.09c

Conversion to km/s = 0.09 * 3.0 * 10^5 km/s = 27000 km/s

Note that this solution, using classic physics do not take into account time and space dilation.

Answer: 27000 km/s

4 0

3 years ago

How can you be both at rest and also moving at 100,000 km/h at the same time

4vir4ik [10]

You could be lying completley still on your bed, and all though it seems you are at rest, you are moving along with the earth around the sun and hence are motion. This is why 'being at rest' is more of a relative term. Hope this helps!

7 0

3 years ago

Which of the following is not a source of light? *<br> a) sun<br> b) star<br> c) mirror<br> d)cfl

Vladimir [108]

I think that would be c) mirror because mirrors reflects light and can't create it.

5 0

3 years ago

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