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Nataly_w [17]
2 years ago
5

A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe

r is heated. As the washer's temperature increases, what happens to the hole in the center? The hole in the center will remain the same size. Changes in the hole cannot be determined without know the composition of the metal. The hole in the center of the washer will expand. The hole in the center of the washer will contract.
Physics
1 answer:
jenyasd209 [6]2 years ago
7 0

answer;

The hole in the center of the washer will expand

explanation;

<em>A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? The hole in the center will remain the same size. Changes in the hole cannot be determined without know the composition of the metal. The hole in the center of the washer will expand. The hole in the center of the washer will contract.</em>

this is an example of area expansivity.

coefficient of area expansivity is change in area per area per degree rise in temperature

a=dA/(A*dt)

as the temperature rises , there will be volumetric and area expansivity on the body. volume also increases because of the intermolecular forces of attraction between the molecule is now getting apart.

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Donna's doctor must have diagnosed her with muscle cramps. With the given symptoms above, it is likely that she is experiencing muscle cramps. Muscle cramps most likely happen when the individual experiencing it is lack of fluid intake, causing the muscles to tighten causing pain and for the muscle to contract involuntarily.
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2 years ago
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Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

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3 0
2 years ago
Select the areas that would receive snowfall because of the lake effect.
Studentka2010 [4]

Answer:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The places that would would get snowfall because of the lake effect are Grand Marais, Two Harbors, and Duluth. The reason for this is that these three places are located right on the shores of the Lake Superior. This lake is one of the biggest lakes in the world. It has enormous amount of water in it, having big impact on the regional climate because of that. The water from this lake creates a lot of humidity in the air, and there's a lot of evaporation as well, both causing the formation of clouds, and when it is cold enough, instead of precipitation, the region gets large amounts of snowfall.

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A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average
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(v) = 0 m/s 
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<span>0 = 20 + a(4) 

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2 years ago
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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

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V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

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Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

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2 years ago
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