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2 years ago

10

A pipe contains water at 500,000 Pa above atmospheric pressure. If you patch a 4.00 mm diameter hole in the pipe with a piece of

bubble gum, how much force must the gum be able to withstand?

1 answer:

IRISSAK [1]2 years ago

7 0

Answer: 6.284N

Explanation:

Pressure is the ratio of force exerted to cross sectional area of the material.

Pressure = Force/Area

Pressure = 500,000Pa

Area = Πd²/4 where d is the diameter of the hole.

If d = 4mm = 0.004m

Area = Π×0.004²/4

Area = 1.26×10^-5m²

Force = Pressure×Area

Force = 500,000× 1.26×10^-5

F = 6.284N

The gum must be able to withstand 6.284N force

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VLD [36.1K]

Answer: there are 15 coins of $2 and 18 coins of $5

Explanation:

I will answer in English.

X is the number of $5 coins.

Y is the number of $2 coins.

We have the system of equations:

Y + X = 33

Y*2 + X*5 = 120

first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:

Y = 33 - X.

Then we can replace it in the other equation:

(33 - X)*2 + X*5 = 120

66 - X*2 + X*5 = 120

X*3 = 54

X = 54/3 = 18

and using the equation for Y.

Y = 33 - X = 33 - 18 = 15

So there are 15 coins of $2 and 18 coins of $5

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2 years ago

A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene

Ilia_Sergeevich [38]

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

$K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}$

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax$

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

$hmax =\frac{vo^{2} }{2*g}$

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

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2 years ago

A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at

Olegator [25]

Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to $1 × 10^{-12}$ meter

1 micrometer is equivalent to $1 × 10^{-6}$ meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by $174 × 10^{-12}$ meter we get that the number of sheets will be as follows

$\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719$

Therefore, they are approximately 719 sheets

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2 years ago

At what condition does a body become weightless at equator

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Hope this is helpful <span>Weightlessness

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2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

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2 years ago