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Dominik [7]
3 years ago
6

Please help . Anybody

Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

1: c

2: a

3: b

sorry if I am wrong

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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
castortr0y [4]
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
4 0
3 years ago
Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.
yan [13]

the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or \frac{kQ}{a}

ATQ,

We have density of charge,

λ = \frac{Q}{L}

Where L is the rod's length, in this case the semicircle's length L = πr

Q is the charge on the rod

The potential created at the center by an differential element of charge is:

k = \frac{dQ}{r}

where k is the coulomb's constant

r is the distance from dQ to center of the circle

v = ∫ \frac{K dQ}{a} , Where a = radius, k = 1 / 4πε∘

v =\frac{kQ}{a} or Q ∕ (4πε∘a)

To learn more about potential from the given link

brainly.com/question/25923373

#SPJ4

6 0
2 years ago
Which organisms were first responsible for depleting carbon dioxide from the atmosphere, replacing it with oxygen?
fgiga [73]
C) Cyanobacteria is the answer
6 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST!!!
kumpel [21]

Answer:

72.53 mi/hr

Explanation:

From the question given above, the following data were obtained:

Vertical distance i.e Height (h) = 8.26 m

Horizontal distance (s) = 42.1 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the car to get to the ground.

This can be obtained as follow:

Height (h) = 8.26 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

8.26 = ½ × 9.8 × t²

8.26 = 4.9 × t²

Divide both side by 4.9

t² = 8.26 / 4.9

Take the square root of both side by

t = √(8.26 / 4.9)

t = 1.3 s

Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:

Horizontal distance (s) = 42.1 m

Time (t) = 1.3 s

Horizontal velocity (u) =?

s = ut

42.1 = u × 1.3

Divide both side by 1.3

u = 42.1 / 1.3

u = 32.38 m/s

Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:

1 m/s = 2.24 mi/hr

Therefore,

32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s

32.38 m/s = 72.53 mi/hr

Thus, the car was moving at a speed of

72.53 mi/hr.

7 0
3 years ago
What would a sound wave do if it traveled from a gas to a
NeTakaya

Answer:

speed up

Explanation:

it's sound, not electromagnetic

5 0
3 years ago
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