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Stels [109]
3 years ago
14

Which of the following is a poor conductor of heat?. A. silver. B. aluminum. C. gold. D. copper. E. air. How can you tell the di

fference if its a poor conductore or not?o.o
Physics
2 answers:
tankabanditka [31]3 years ago
8 0

Definitely air since most metals are conductors and air is an insulator.

Rama09 [41]3 years ago
3 0
The answer is E. air
Air is a poor conductor of heat. This can be determined through a property called thermal conductivity and has a unit of W/m.K. It is described as a quantity of heat through a unit thickness of material per temperature. The higher the thermal conductivity of a material, the more it is able to conduct heat. Most gases are poor conductors of heat.
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HELP! FOR A TEST!
Sati [7]

Answer:

<h3>On a velocity vs time graph the slope of the line represents the acceleration of the object. With a slope of zero, the object is moving at a constant velocity in the positive (+) direction during this five minute interval. ... Displacement and distance can both be determine on a velocity vs.</h3>

4 0
3 years ago
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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that
Harrizon [31]

Answer:

The time taken is   t = 40007 sec  

Explanation:

From the question we are told that

   The diameter of the egg is d_e = 5.5cm  = \frac{5.5}{100} = 5.5*10^{-2}m

    The initial temperature of egg the T_e = 4.3^{o}C

     The temperature of the boiling water T_b = 100^oC

    The heat transfer coefficient is  H  = 800 W/m^2 \cdot K

    The  final temperature is T_e_f = 74^oC

     The  thermal  conductivity of water is k = 0.607 W/m^oC

     The diffusivity of the egg \alpha = 0.146 * 10^{-6} m^2 /s

Using one term approximation

We have the

            \frac{T_e_f - T_b}{T_e - T_b}  = Ae^{-\lambda ^2 \tau}

The radius is  r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of A  \ and \ \lambda to use in the above equation

     The Biot number is mathematically represented as

               Bi = \frac{H r}{k}

Substituting values  

               Bi = \frac{800 * 2.75 *10^{-2}}{0.607}

                    = 36.24

So for this value  which greater than 0.1 the  coefficient \lambda_1 \ and  \ A_1 is  

        \lambda = 3.06632

        A = 1.9942

Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

           0.13624 =  e^{-9.383 \tau}

Taking natural log of both sides

           -1.993 =  -9.383\  \tau

          \tau =  0.2124

    The time required for the egg to be cooked is  mathematically represented as

          t = \frac{\tau r^2}{\alpha }

substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

         t = 40007 sec  

8 0
3 years ago
A train is moving at a speed of 50 km/h. How many hours will it take the train to travel 600 kilometers?
EastWind [94]

Answer:

12 hours

Explanation:

600 divided by 50 is 12

6 0
3 years ago
Read 2 more answers
How many times does the earth spin in a year
sp2606 [1]
It spins one bc it takes 365 days to make a year and to make a year the earth has to mov some
7 0
3 years ago
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A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

5 0
3 years ago
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