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2 years ago

Which properties of a lava lamp are important to remember when modeling convection?

1 answer:

5 0

Important thing when making a lava lamp are the two liquids with different density so that it would not mix and that it would expand when heated. The movement of the orange liquid is called convection. if you will not heat the orange one, it till just stay on the stop and not go under tha water.

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What are some of the physical properties of stars?

Arisa [49]

Distance, luminosity, brightness, radius, chemical composition and temperature

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2 years ago

How much water

nevsk [136]

Answer:

The mass is $m = 3.75 \ kg$

Explanation:

From the question we are told that

The initial temperature is $T_1 = 15^oC$

The final temperature is $T_2 = 100 ^o C \ (boiling point )$

Generally the maximum heat produced by 1 Liter of natural gas is $9000 \ cal$

So the amount of heat produced by 100 L is

$E = 9000 * 100$

=> $E = 9000 00 \ cal$

Generally given that the efficiency is $\eta = 0.35$

Then actual heat received by the water is

$H = 0.35 * E$

=> $H = 0.35 * 9000 00$

=> $H = 315000 \ cal$

Converting to kcal

=> $H = 315000 \ cal = \frac{315000}{1000} = 315 \ kcal$

Generally the specific heat of water is

$c_w = 1 kcal/ kg \cdot ^oC$

Generally the heat received by the water is mathematically represented as

$H = m * c_w * (T_2 - T_1)$

=> $315 = m * 1 * ( 100 - 15 )$

=> $m = 3.75 \ kg$

6 0

2 years ago

Which of the following are vector quantities? Check all that apply.

Dennis_Churaev [7]

It’s distance for my answer but tell me if anyone tells u this answer thanks

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2 years ago

An astronomer would most likely use the doppler effect to question 12 options: 1) measure the color of a star. 2) measure the pa

zalisa [80]

The answer would be number four. I'm sorry if I am too late. Byes.....

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2 years ago

Read 2 more answers

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$ , the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q = -3.923\,C$ and $r = 100\,m$ , then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0

2 years ago