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Len [333]
3 years ago
7

How to create a water ripple experiment

Physics
1 answer:
Phoenix [80]3 years ago
5 0
I once had to create a water ripple experiment. I took a tote covered the bottom with rocks then filled the tote half way full with water. i made a ramp out of a shoebox or similar and took a bouncy ball and rolled it down the ramp. The thing I should have done is have something so you could tell how strong the wave was.

I don't know if your teacher would take this When I did it we were talking about natural disasters so ask your teacher first!
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A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering
Black_prince [1.1K]
<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y is the height of the stone at 6s (the value we want to find)

y_{o}=925ft is the initial height of the stone

V_{o}=20ft/s is the initial velocity of the stone

t=6s is the time  at which we need to find the height

g=32ft/s^{2} is the acceleration due to gravity

Having this clear, let's find y from (1):

y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2} (2)

Finally:

y=469ft This is the height of the stone at t=6s

4 0
3 years ago
How much work is done when 0.0080 C is moved through a potential difference of 1.5 V?
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.012 J is going to be your answer

3 0
3 years ago
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BRAINLIEST. Agraph is probelow. The graph shows the speed of a car traveling east over a 12 second period. Based on the informat
gavmur [86]

Answer: Speeding up constantly

Explanation:

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Which type of cloud would you expect to be involved in some form of precipitation?
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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

5 0
3 years ago
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