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otez555 [7]
2 years ago
8

Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

ed of the water waves? What is the period of these water waves?
Physics
1 answer:
snow_lady [41]2 years ago
4 0

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength \lambda=0.06m.

The frequency f of the waves is 14.8 waves every 3 seconds or

f=14.8/3 =4.33Hz.

Now the relationship between wavelength \lambda, frequency f and speed v of the waves is:

v=\lambda f

We put in the values \lambda=0.06m and f=4.933Hz and get:

\boxed{v=0.06*4.922=0.296m/s}

Now the period T is just the inverse of the frequency, or

T=\frac{1}{f}

\boxed{T=\frac{1}{4.933}=0.203\:seconds }

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A horizontal pipe of diameter 0.81 m has a smooth constriction to a section of diameter 0.486 m . The density of oil flowing in
Lerok [7]

Answer:

2.06 m³/s

Explanation:

diameter of pipe, d = 0.81 m

diameter of constriction, d' = 0.486 m

radius, r = 0.405 m

r' = 0.243 m

density of oil, ρ = 821 kg/m³

Pressure in the pipe, P = 7970 N/m²

Pressure at the constriction, P' = 5977.5 N/m²

Let v and v' is the velocity of fluid in the pipe and at the constriction.

By use of the equation of continuity

A x v = A' x v'

r² x v = r'² x v'

0.405 x 0.405 x v = 0.243 x 0.243 x v'

v = 0.36 v' .... (1)

Use of Bernoulli's theorem

P+\frac{1}{2} \rho v^{2}=P' +\frac{1}{2}\rho'v'^{2}

7970 + 0.5 x 821 x 0.36 x 0.36 x v'² = 5977.5 + 0.5 x 821 x v'²    from (i)

1992.5 = 357.3 v'²

v' = 5.58 m/s

v = 0.36 x 5.58

v = 2 m/s

Rate of flow = A x v = 3.14 x 0.405 x 0.405 x 2 x 2 = 2.06 m³/s

Thus the rate of flow of volume is 2.06 m³/s.

5 0
2 years ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
Sunny_sXe [5.5K]

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

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\hat B = \hat v \times \hat r

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\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

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So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

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2 years ago
A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi
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Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
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Equation:
v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \:  \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142
we \: can \: only \: assume \:that \\  flow \: (q) \:stays \: same \: since \: it \\  isnt \: impeded \: by \:  anything \\ thus \:it  \: (q)\:  stays \: the \: same \:  \\ so \: 4q \: can \: be \: removed \: from \:  \\ the \: equation
then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to >  >  \\ (v1 \times {d1}^{2} \pi) = (v2  \times   {d2}^{2}\pi)
both \: \pi \: will \: cancel \: each \: other \: out \:  \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \:  =

(v1  \times   {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times   {0.052}^{2}) = (v2  \times   {0.021}^{2}) \\ divide \: both \: sides \: by \:  {0.021}^{2} \\ to \: solve \: for \: v2 >  >
v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2})  \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second
new velocity coming out of the hose then is
44 ft/sec
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