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Oksana_A [137]
4 years ago
10

How will heat flow between objects?

Physics
2 answers:
zzz [600]4 years ago
5 0
Without an external agent doing work,heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There arethree different ways for heat to flowfrom one object to another. They areconduction, convection, and radiation



How hope this helps❤
Pachacha [2.7K]4 years ago
3 0

Question: How will heat flow between objects?

from the colder object to the warmer object

from the bottom object to the object on top

from the warmer object to the colder object

back and forth between objects of equal temperature

Answer: Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.

Conduction: The process by which heat or electricity is directly transmitted through a substance when there is a difference of temperature or of electrical potential between adjoining regions, without movement of the material.

Convection: The movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat.

Radiation: The emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles that cause ionization.

Hope this helps, have a good day. c;

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It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atm
klio [65]

Answer:

a) mb = 0.0596 kg ; r = 0.974 m

b) a = 754 m/s^2 .. (Upward)

c) mL = 5.96 kg

Explanation:

Given:-

- The density of Mars atmosphere , ρ = 0.0154 kg/m^3

- The surface density of ballon, σ = 5.0g/m^2

Solution:-

(a) What should be the radius and mass of these balloons so they just hover above the surface of Mars?

- We will first isolate a balloon in the Mar's atmosphere and consider the forces acting on the balloon. We have two forces acting on the balloon.

- The weight of the balloon - "W" - i.e ( Tough plastic weight + Gas inside balloon). Since, the balloon is filled with a very light gas we will assume the weight due to gas inside to be negligible. So we have:

                            W = mb*g

Where,  mb: Mass of balloon

             g: Gravitational constant for Mars

- The mass of the balloon can be determined by using the surface density of the tough plastic given as "σ" and assuming the balloon takes a spherical shape when inflated with surface area "As".

                           As = 4πr^2

Where,  r: The radius of balloon

So,                      mb =  4σπr^2

- Substitute the mass of balloon "mb" in the expression developed for weight of the balloon:

                         W = 4*σ*g*πr^2    ......... Eq1

- The weight of the balloon is combated by the buoyant force - "Fb" produced by the volume of Mars atmosphere displaced by the balloon acting in the upward direction:

                        Fb = ρ*Vs*g

Where,    Vs : Volume of sphere = 4/3 πr^3

So,                    Fb = ρ*g*4/3 πr^3   ....... Eq 2        

- Apply the Newton's equilibrium conditions on the balloon in the vertical direction:

                       Fb - W = 0

                       Fb = W

                       ρ*g*4/3 πr^3 = 4*σ*g*πr^2        

                       r = 3σ / ρ

                       r = 3*0.005 / 0.0154

                       r = 0.974 m           .... Answer            

- Use the value of radius "r" and compute the "mb":

                       mb =  4σπr^2

                       mb =  4*0.005*π (0.974)^2  

                       mb = 0.0596 kg   ... Answer  

(b) If we released one of the balloons from part (a) on earth, where the atmospheric density ρ = 1.20kg/m^3, what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down?

- The similar analysis is to be applied when the balloon of the same size i.e r = 0.974 m and mass mb = 0.0596 kg is inflated on earth with density  ρ = 1.20kg/m^3.

- Now see that the buoyant force acting on the balloon due to earth's atmosphere is different from that found on Mars. So the new buoyant force Fb using Eq2 is:

                       Fb = ρ*g*4/3 πr^3

Where,   g: Gravitational constant on earth = 9.81 m/s^2

                       Fb = (1.20)*(9.81)*(4/3)* π*(0.974)^3

                       Fb = 45.5 N

- Apply the Newton's second law of motion in the vertical direction on the balloon:

                      Fb - W = mb*a

Where,          a: The acceleration of balloon

                     a = (Fb - W) / mb

                     a = Fb/mb - g

                     a = 45.5/0.0596 - 9.81

                    a = 754 m/s^2  (upward) ..... Answer

c), d) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

- The new radius of the balloon - "R" -is five times what was calculated in part (a):

- Apply the Newton's equilibrium conditions in the vertical direction on the balloon with the addition of downward weight of load "WL":

                     Fb - W - WL = 0

                     WL = Fb - W

                     mL*g = ρ*g*4/3 πR^3 - 4*σ*g*πR^2      

Where,          mL : The mass of load due to instrument package

                     mL =  ρ*4/3 πR^3 - 4*σ*πR^2

                     mL = 0.0154*4/3*π*(5*0.974)^3 - 4*(0.005)*π*(5*0.974)^2    

                     mL = 7.45 - 1.45

                     mL = 5.96 kg   ..... Answer                      

6 0
3 years ago
Read 2 more answers
Question 1 of 5
liubo4ka [24]

Answer:

D

Explanation:

its mass would be the same

because the mass of an object is constant or its independent of the gravitational force or pull

5 0
3 years ago
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