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nika2105 [10]
3 years ago
10

Suppose the ends of a 27-m-long steel beam are rigidly clamped at 0°C to prevent expansion. The rail has a cross-sectional area

of 35 cm^2. What force does the beam exert when it is heated to 39°C? (αsteel = 1.1 × 10^−5/C°, Ysteel = 2.0 × 10^11 N/m^2).
Physics
1 answer:
frozen [14]3 years ago
5 0

Answer:

F = 1.58*10^{11} N

Explanation:

given data:

length of steel beam = 27 m

cross sectional area of rail = 35 cm

\Delta T = 39 Degree celcius

change in length of steel beam is given as

\Delta L = L_O \alpha \Delta T

            = 20*1.1*10^{-5}*39

           =8.58*10^{-3} m

Young's modulus is

Y = \frac{FL}{A\Delta L}

F = \frac{ YA\Delta L}{L}

= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}

F = 1.58*10^{11} N

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The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>
  • We have the Poiseuille's formula,

                     Q=\frac{\pi r^4P}{8\beta l}

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  • By substituting values and rearranging we will get the pressure drop as,

                  P=3.284Pascals

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Learn more about the Poiseuille's formula here:

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What is force of gravity​
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13. Which statement correctly describes the human heart? A. It has three chambers, and pumps blood directly into veins. B. It ha
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What are the subzones for each main zone?
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Answer:

The oceanic zone is subdivided into the epipelagic, mesopelagic, and bathypelagic zones on the basis of amount of light that reaches different depths. The mesopelagic (disphotic) zone, where only small amounts of light penetrate, lies below the Epipelagic zone.

Explanation:

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3 years ago
A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a ne
blagie [28]

Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

3 0
3 years ago
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