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Evgesh-ka [11]
3 years ago
5

At about 55 meters/sec, a falling parachuter (before the parachute opens) no longer accelerates. Air friction opposes accelerati

on. Although the affect of air friction begins gradually, imagine that the parachuter is free falling until terminal speed ( the constant falling speed) is reached. How long would that take?
Physics
1 answer:
JulijaS [17]3 years ago
3 0
The acceleration of gravity on Earth is about 9.8 m/s².
That means that if air resistance is neglected, a falling body
smoothly gains 9.8 m/s of downward speed each second.

If our intrepid chutist lightly stepped out of the plane or was
gently pushed, so that his initial downward speed was zero
and grew by 9.8 m/s every second, then it took him ...

               (55) / (9.8)  =  5.6 seconds

to reach that particular downward speed.
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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
At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.
tatuchka [14]
<span>At the periphery of a hurricane the air is sinking, and several kilometers above the surface, in the eye, the air is sinking.  </span>
6 0
3 years ago
Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi
Vedmedyk [2.9K]

Answer:

v = 1.98\ m/s

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m             u₄ = 4.5 m/s

m₅ = m              u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

v = \dfrac{9.9}{5}

v = 1.98\ m/s

5 0
3 years ago
If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

3 0
3 years ago
Group elements number 11 to 20 as either metallic, non metallic or metalloid.​
bija089 [108]

Answer:

The elements are grouped into the different substances by color. As you can see, Lithium, Beryllium, Sodium, Magnesium, Aluminum, Potassium, and Calcium are metals out of the first 20 elements.

Hydrogen, Helium, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Phosphorus, Sulfur, Chlorine, and Argon, are non-metals within the first 20 elements.

Boron and Silicon count as Metalloids in the Periodic Table (properties of both metals and non-metals)

reference- socatric q and a

Explanation:

7 0
3 years ago
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