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lina2011 [118]
3 years ago
14

O professor Hosney, levou os alunos da segunda série ao laboratório para realizar um experimento. Pegou um recipiente de capacid

ade 1000ml, a uma temperatura de 68oF e nele despejou 980 ml de uma substância, a 20oC. Enquanto colocou o conjunto para aquecer, consultou uma tabela onde encontrou o coeficiente de dilatação volumétrica da substância, 4 x 10-4 ºC-1 e o coeficiente de dilatação linear do material do recipiente, 3 x 10-5 ºC-1. Hosney pediu então que os alunos determinassem a temperatura a partir da qual, a substância iria transbordar. Um aluno perguntou então, qual a temperatura de fusão da substância, e o professor respondeu prontamente 290,8 K. Qual a temperatura a partir da qual a substância transbordará ?
Physics
1 answer:
givi [52]3 years ago
8 0

Answer:

The temperature beyond which the substance overflows the container is 86.23°C.

Explanation:

English Translation

Professor Hosney took the second grade students to the laboratory to perform an experiment. He took a 1000ml capacity container at a temperature of 68oF and poured 980 ml of a substance at 20oC into it. While placing the set to heat, he consulted a table where he found the volumetric expansion coefficient of the substance, 4 x 10-4 ºC-1 and the linear expansion coefficient of the container material, 3 x 10-5 ºC-1. Hosney then asked students to determine the temperature from which the substance would overflow. A student then asked, what is the melting temperature of the substance, and the teacher answered promptly 290.8 K. What is the temperature from which the substance will overflow?

Solution

The change in volume of a substance is given as

ΔV = γV₀(ΔT)

where

γ = coefficient of volume expansion

V₀ = Initial volume

(ΔT) = change in temperature.

At the temperature where the substance will overflow, the volume of the substance and the container will both be the same.

Let this temperature be T.

For the substance,

γ = coefficient of volume expansion = (4 × 10⁻⁴) °C⁻¹

V₀ = Initial volume = 980 mL

(ΔT) = change in temperature = (T - 20)

We will still leave ΔT as ΔT

ΔV₁ = (4 × 10⁻⁴) × 980 × ΔT

ΔV₁ = 0.392 ΔT

New volume of the substance at that temperature = V₀ + ΔV₁ = 980 + 0.392ΔT

For the container

γ = coefficient of volume expansion = 3 × coefficient of linear expansion = 3 × (3 × 10⁻⁵) °C⁻¹ = (9 × 10⁻⁵) °C⁻¹

V₀ = Initial volume = 1000 mL

(ΔT) = change in temperature = (T - 20) (note that 68°F = 20°C)

We will still leave ΔT as ΔT

ΔV₂ = (9 × 10⁻⁵) × 1000 × ΔT

ΔV₂ = 0.09 ΔT

New volume of the container at that temperature = V₀ + ΔV₂ = 1000 + 0.09 ΔT

At the temperature where overflow occurs, the two volumes are initially first the same.

980 + 0.392ΔT = 1000 + 0.09 ΔT

0.392ΔT - 0.09ΔT = 1000 - 980

0.302ΔT = 20

ΔT = (20/0.302) = 66.23°C

T - 20° = 66.23°

T = 66.23 + 20 = 86.23°C

The temperature beyond which the substance overflows the container is 86.23°C.

Hope this Helps!!!

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SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0
3 years ago
According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind
Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0
3 years ago
A soccer player kicks a soccer ball initially at rest setting it in motion at a velocity of 30 m/s. If the ball has a mass of 0.
Alchen [17]

Answer:

 F= 600 N

Explanation:

Given that

Initial velocity ,u= 0 m/s

Final velocity ,v= 30 m/s

mass ,m = 0.5 kg

time ,t= 0.025 s

The change in the linear momentum is given as

ΔP= m (v - u)

ΔP= 0.5 ( 30 - 0 ) kg.m/s

ΔP= 15 kg.m/s

We know that from second law of Newtons

F=\dfrac{dP}{dt}

F=\dfrac{\Delta P}{t}

Now by putting the values

F=\dfrac{15}{0.025}\ N

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5 0
3 years ago
How much heat is absorbed by 41 g iron skillet when its temperature varies from 9°C. to 20°C
alexgriva [62]

Answer:

202.95J

Explanation:

The formulae for the energy absorbed by the iron skillet is;

q=m*c*ΔФ   where q=energy absorbed, m is mass of iron skillet, c is specific heat capacity and ΔФ is change in temperature

Given that;

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c=0.45 J/g°C

q= 41×0.45×11

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4 0
3 years ago
A wedge shaped air film is made between two sheets of glass using a spacer at one end of the glass sheets. If light of wavelengt
pychu [463]

Answer:

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Explanation:

We are given that

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1nm=10^{-9} m

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Suppose that first dark fringe and fifth dark fringe near spacer, then the path length of light is 4 times the wavelength of light.

The light passes through air film is two times  then the change in air film thickness  from one edge to other is two times the wavelength of light.

Change in air film thickness  from one edge to other edge  is same as the thickness of spacer.

Therefore, thickness of spacer=2\lambda

Thickness of spacer=2\times 589\times 10^{-9}m

Thickness of spacer=1178 nm

Hence, the thickness of spacer=1178 nm

8 0
2 years ago
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