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PolarNik [594]
2 years ago
9

Gaya adalah...?? [tolong yaaa..]

Physics
1 answer:
ludmilkaskok [199]2 years ago
8 0

I can try to help if you speak English

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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
In example 20.3 in the text, the net force on a 1.0 nC charge located between two 10 nC charges is calculated. How would the ans
VashaNatasha [74]

Answer:

The forces experienced by the middle particle are attractive, and the net force will remain the same (0) if and only if the distances of the sides particles to the middle particle are the same.

Explanation:

In example 20.3 the forces experienced by the middle particle are repulsive because all the particles are positive, for the case in which the particles on the sides are replaced for negative charge particles the forces experienced by the middle particle are attractive. Regarding the net force, because we don't know the distances we can not give a definitive answer, what we can say is that if the distances from the middle particle to the sides particles are the same the net force is zero for both cases (remain unchanged).  

7 0
4 years ago
The velocity of a projectile at launch has a
arlik [135]

Answer:

1. 0 vh g

Explanation:

As air resistance is negligible, horizontal speaking, nothing is affecting the velocity. So the horizontal velocity of the projectile stays the same, vh.

As for vertical velocity, since there's always a constant gravitational acceleration acting downward, namely g, the vertical speed will decrease until it reaches the top where it is 0, it then starting to increase in magnitude, downward, due to its gravitational acceleration g.

7 0
3 years ago
The outside diameter of your teacher's rear bicycle tire is 16 inches. How far will he travel if the rear wheel makes 1200 revol
FinnZ [79.3K]

Answer:

241,274.32 inches

Explanation:

How far will he travel if the rear wheel makes 1200 revolutions on the road?

Since the rear wheel makes one revolution in the distance of a circumference of a circle, C with diameter, d = 16 inches

C = πd²/4

So, the distance, travelled in 1200 revolutions is D = 1200 × C = 1200πd²/4

Substituting d = 16 into D, we have

D = 1200πd²/4

D = 1200π(16)²/4

D = 76800π

D = 241,274.32 inches

3 0
3 years ago
A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of
Aneli [31]

Explanation:

Let's solve the problem by using conservation of momentum:

m1 u1 + m2 u2 = m1 v1 + m2 v2

where:

m1 = 3.5 g  is the mass of the blue marble

m2= 1.2 g is the mass of the red marble

u1 = 15 cm/s  is the initial velocity of the blue marble

u2 = 3.5 cm/s is the initial velocity of the red marble

v1 = 5.5 cm/s is the final velocity of the blue marble

We can find the final velocity of the red marble by re-arranging the equation and solving for v2 :

v2 = 1/m²(m1 u1 + m2 u2 - m1 v1)=

=1/1.2(3.5×15+1.2×3.5−3.5×5.5 )=31cm/s

7 0
4 years ago
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