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2 years ago

If we pour very hot liquid like tea or coffee

1 answer:

7 0

Answer: Hello, I dont know if this gives much information but here. Glass comes in the insulator class of material which means it is a poor conductor of heat. when you pour a very hot liquid into a thick tumbler what happens is, the surface of glass that comes in contact with the hot liquid heats up and expands as per its coefficient of thermal expansion, while the outer layer still remain cold and does not expands (because glass is a poor conductor and does not carry heat quickly). This leads to development of thermal stresses which eventually causes cracking.

Explanation:

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago

How are wave properties and energy related? give me examples

klasskru [66]

Answer:

They don’t ‘represent’ anything, they are properties of the wave.

Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.

If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.

For many waves, we don’t perceive them at all (e.g. radio waves).

For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.

6 0

2 years ago

3. Write any two types of force.

Lerok [7]

Answer:

An applied force is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.

A friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:

$F_{frict}$ = µ • $F_{norm}$

8 0

2 years ago

Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov

makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

Calculate the integral then divide the result by 16.

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = 2.021

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

Calculate the integral then divide the result by 16.

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = 0.3542

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

Calculate the integral then divide the result by 16.

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = 0.6875

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0

2 years ago

_____ has a longer wavelength than _____. _____ has a longer wavelength than _____. Blue ... green Green ... yellow Red ... gree

nadya68 [22]

Answer:

red has a longer wavelength than yellow. Yellow has a longer wavelength than green.

Explanation

in the visible spectrum a color with high frequency contains shorter wavelength.In this case red color has high wavelength followed by yellow followed by green,blue and violet.

3 0

3 years ago

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