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3 years ago

If a spear is jabbeb exactly at that place where the fish appears in water, the fish is not killed

Physics

1 answer:

Korvikt [17]3 years ago

7 0

You have learned your lesson well, Suhay. Your statement is correct.

The light rays from the fish BEND when they flow out of the water into the air. But our primitive brain still believes that the light rays flow STRAIGHT from the fish. The result is that the fish does not APPEAR to be at that place where it really is.

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The table below shows the measurements you took in an experiment. Trial Length ( miles) 1.9 4.2 N 3 5.9 4 What is the longest me

spin [16.1K]

Answer:

9.5km

Explanation: just got it right

7 0

3 years ago

Higher-frequency sound waves are perceived as having _____.

Alona [7]

Higher frequency sound waves are perceived as having higher pitch

7 0

3 years ago

A point source at the origin emits sound of frequency 175 Hz uniformly in all directions. On the x-axis at x=100 m, the sound in

hammer [34]

Answer:

Multiple answers:

1. Power output P=17.59W

2.Intensity 160m I=17.6W/ $m^{2}$

3. dB = 77.3

4. f=178.5 Hz

Explanation:

First one comes from the expression

$I=\frac{P}{4\pi r^{2} }$

where I is the intensity, P is the power and r is the radio of the spherical wave, or in this case, the distance x. I solved for the Power by multiplying Intensity with the area (4 $\pi x^{2}$

Second one is done with:

$\frac{I_{2} }{I_{1} } =\frac{x^{2}_{1} }{x^{2} _{2}}$

Solving for Intensity 2, the result mentioned.

The third is simply computed with

$dB=10*log\frac{I}{10^{-12} }$

And finally the last one is done with doppler effect, taking into account the speed of the air as in 10ºC 337m/s.

$f=f_{initial} *(\frac{s+v_{receiver} }{s+v_{source} } )$

Where Finitial is the frequency emitted and s is the speed of the sound. The wind blowing in positive is, in principle, going away of the observer.

8 0

3 years ago

Why is it when a ballerina stands on her toes she exerts more pressure than an elephant on all fours

Murrr4er [49]

I'm not sure that's true ... we'd need to see some numbers.
But if it is true, it's the same reason why some places don't
want ladies to walk on their floors with high heels.

Pressure is (weight) divided by (area).

Pressure exerted
by a ballerina = (ballerina's weight)/(area of her toe on the floor)

Pressure exerted
by an elephant = (elephant's weight)/(area of 4 elephant feet)

Pressure exerted by
a lady in high heels = (lady's weight) / (area of 1 or 2 heels)

You can see that the greatest weight might not exert the greatest
pressure, and the smallest weight might not exert the smallest
pressure. The area is important, and it takes an awful lot of toes
or high heels to equal the area of one or more elephant's feet.

7 0

4 years ago

Al is floating freely in her spacecraft, and you are accelerating away from her with an acceleration of 1g. 5) How will you feel

wariber [46]

Answer:

D. You will feel the same weight as you do on Earth

Explanation:

In free space, she is suppose to be weightless.

Free fall can be described as body in motion where the body is under the effect of acceleration due to gravity only and no other acceleration..

Since I am accelerating away from her at an acceleration of 1g

Then,

F=ma, where a=g

Then F=mg

Since my weight on earth is W=mg

This is equals to my weight in the spaceship, then I will feel the same weights as I do on earth.

5 0

4 years ago