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3 years ago

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Water is pumped through a pipe of diameter 15.0 cm, from the Colorado River up to Grand Canyon Village, located on the rim of th

e canyon. The river is at an elevation of 564 m, and the village is at an elevation of 2096 m. What is the minimum pressure at which the water must be pumped if it is to arrive at the village?

1 answer:

Aleksandr [31]3 years ago

8 0

Answer:

p= 1.50289×10⁷ N/m²

Explanation:

Given

HA = (564 m)................(River Elevation)

HB = (2096 m).............(Village Elevation)

Area = A =(π/4){Diameter}² = (π/4){0.15 m}² = 0.017671 m²

ρ = (1 gram/cm³) = (1000 kg/m³)........(Water Density)

p(pressure)=?

Solution

p=PA - PB

p= ρ*g*HB - ρ*g*HA

p= (ρ*g)*(HB - HA)

p= (1000×9.81 )×{2096 - 564}

p= 1.50289×10⁷ N/m²

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Bogdan [553]

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t $_{min$ = ( m + 1/2) × λ/2n

we substitute

t $_{min$ = ( 0 + 1/2) × 711 /2(1.21)

t $_{min$ = 0.5 × 711/2.42

t $_{min$ = 0.5 × 293.80165

t $_{min$ = 146.9 nm

Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm

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3 years ago

Carbon-14 is used to determine the time an organism was living. The amount of carbon-14 an organism has is constant with the atm

lutik1710 [3]

Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

$\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} }$ (1)

Where:

$n_{o}$ - Initial amount of carbon-14.

$n(t)$ - Current amount of carbon-14.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

Then, we clear the time within the formula:

$t = -\tau \cdot \ln \frac{n(t)}{n_{o}}$ (2)

In addition, time constant can be calculated by means of half-life of carbon-14 ( $t_{1/2}$ ), measured in years:

$\tau = \frac{t_{1/2}}{\ln 2}$

If we know that $\frac{n(t)}{n_{o}} = 0.25$ and $t_{1/2} = 5730\,yr$ , then the age of the organism is:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.25$

$t \approx 11460.001\,yr$

The age of the organism is approximately 11460 years.

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3 years ago

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When the distance between two interacting objects doubles, the gravitational force is

Umnica [9.8K]

The gravitational force will be one quarter.

The gravitational force between two objects is given by the formula

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Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.

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3 years ago

This. please help me.

kramer

The transfer of heat from the Sun is in the form of thermal radiation (also called infrared radiation)
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The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th

jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

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P₁ = 130 psig

i.e.,

= $130\times 6.894$

= $896.22 \ Kpa$

or,

= $896.22\times 10^3 \ Pa$

Z₂ = 10ft

= 3.05 m

$\delta$ = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒ $\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2$

On substituting the values, we get

⇒ $\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2$

⇒ $\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05$

⇒ $P_2=866330 \ P_a$

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