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atroni [7]
2 years ago
7

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

th a speed of 7.0 feet per second. If you then put your thumb over the end of the hose so the opening is only a circle 0.25" across, with what speed will the water stream out now?
Physics
1 answer:
castortr0y [4]2 years ago
4 0
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
2) 0.25in × 1ft/12in = 0.021ft
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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Answer:

halved

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Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti
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A) 1.36\cdot 10^{-4}T

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Here we have

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B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

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C) 3.26\cdot 10^{-23}N

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F=qvB

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2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
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Answer:

Explanation:

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It's melting point is 1336 K

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The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

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2 years ago
A negatively charged object is located in a region of space where the electric field is uniform and points due north. the object
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\Delta U = e d

- The second largest increase is when the charge is moving east. In this case, actually, the variation of potential energy is zero. This is because the charge is moving perpendicular to the field, and so it is moving along points with same potential. Therefore, in this case the variation of potential energy is zero:
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Yuki888 [10]

Answer:

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(3)T=(0)*1/2=0

So the system is still balanced

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2 years ago
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