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poizon [28]
3 years ago
13

Leila is building an aluminum-roofed shed in her backyard to store her garden tools.The flat roof will measure 2.0 x 3.0m in are

a during the coldest winter months when the temperature is -10° C, but temperatures in Leila’s neighborhood can reach as high as 38° C in the summer. What is the area of the roof that should stick out from the shed in the summer so that the roof just fits the structure during cold winter nights? (aluminum= 24 x
Physics
1 answer:
IrinaK [193]3 years ago
3 0
Answer: 0.0138 m^2  = 138 cm^2

Explanation:

The thermal expansion is the term use for the physical phenomena of dilation of the objects when they are exposed to changes in temperature.

The objects dilate when they are heated and contract when they are cooled.

The dilation is proportional to the change in temperatur.

For linear dilation, the proportionality constant is called linear dilation coefficient of the materials, it is named α and is measured in °C ^-1.

ΔL = α * Lo * ΔT, which means that the dilation (or contraction) is proportional to the product of the original length (Lo) and the change of temperature (ΔT).

There is also superficial dilation, for which the dilation is:

ΔA = β * Ao * ΔT, which means that the superficial dilation (or contraction) is proportional to the product of the original area (Ao) and the change of temperature (ΔT).

It is very interesting and important to solve problems that β = 2α, because regularly you will find the values of α for different materials and so, you just to multiply it times 2 to use β.

For this problem:

- Original area, Ao = area of the flat roof at - 10°C = 2.0m * 3.0m = 6.0 m^2.
- α for aluminum = 24 * 10^ -6 °C^-1.
- ΔT = 38°C - (-10°C) = 48°C

So, ΔA = 6.0m^2 * (2 * 24*10^-6 °C&-1) * 48°C = 0.0138 m^2

And that is the area that should stick out in summer to fit the structure during cold winter nights.

You can pass that number to cm^2 to grasp better the idea of this size:

0.0138 m^2 * (100 cm)^2 / m^2 = 138 cm^2




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One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
3 years ago
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and i
zavuch27 [327]

Answer:

ΔL = 1.011 mm

Explanation:

Let's begin by listing out the given information:

Length (L) = 600 mm = 0.6 m,

Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,

Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,

Modulus of Elasticity (E) = 85 GN/m²,

Compressive load (F) = 180 KN

Using the formula, Stress = Load ÷ Area

Mathematically,

σ = F ÷ A = 180 x 10³ ÷ 0.001256

σ = <u>143312.1 KN/m</u>²

Modulus of elasticity = stress ÷ strain

E = σ ÷ ε

ε = ΔL/L

85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)

ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶

ΔL = L x 1686.02 * 10⁻⁶

ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶

ΔL = 1.011 x 10⁻³ m

ΔL = <u>1.011 mm</u>

∴The bar contracts by 1.011 mm

4 0
3 years ago
Plan an experiment to measure the ideal mechanical advantage of a three-hole punch. (a) What materials would you need? (b) What
Vaselesa [24]

Answer:

A) Three hole punch and either a layered plastic or paper

B) Identify the lengths involved  ,

  Length of input arm / length of output arm = L1/ L2

Explanation:

<u>a) Materials involved includes :</u>

Three hole punch and either a layered plastic or paper

Identify the forces acting on the three-hole punch which are Input and output forces

Identify the points where they act

<u>B) procedures involved </u>

The mechanical advantage = output force / input force

step one:  Identify the lengths involved

assuming no friction or relatively small friction \

mechanical advantage can be calculated as : Length of input arm / length of output arm = L1/ L2

7 0
3 years ago
Water in the air combines with
valentina_108 [34]

Answer:b

Explanation:

6 0
3 years ago
A student combined equal amounts of two solutions. One solution had a pH of 2 and the other had a pH of 12. Which would most lik
zalisa [80]
Answer: optio C. 6.

pH measures the acidity of the solutions.

pH 2 is highly acid and pH 12 is highly basic (alkalyne).

You can be sure that the pH of the solution will be intermediate between 2 and 12.

Given that the volumen of the two solutions are equal the pH of the resulting combination cannot be so close to the original solutions. That takes you to eliminate all the options except the option C. 6.

You cannot be sure that the final pH will be 6, because that will depend on the stregths of the two original solutions, that is why the question asks for the most likely resulting pH, and the answer is optio C. 6. 
6 0
3 years ago
Read 2 more answers
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