Describe the difference between a conductor and an insulator in terms of thermal energy transfer and provide an example of each.

Answers

Answer 1

Answer:

The difference between conductors and insulators is because of electronic structure of atoms and molecules.

Anywhere we have free electrons, we have a hard conductor where free electrons help in conduction.

In electrolytes, charged Ions do conduction.

Dielectrics (or insulators) don't have free electrons current.

Free electrons mean, the electrons that are not related with a specific atom, they are allowed to move nearby the crystal lattice.

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When Tito lifts the heaviest object he can, he is testing his _____.
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Holden is trying to determine the velocity of his race car. He went 20 meters east, turned around, and went 40 meters west. He turned the car one more time and went 35 meters east. His car was 15 meters from the starting line. This took 5 seconds.What is the car’s velocity?
A. 3 m/s west

B.3 m/s east

C. 19 m/s west

D. 19 m/s east

Answers

Although the car's average speed is 19 m/s ,
its velocity is 3 m/s east.

Velocity is (displacement) / (time)

and Displacement is the distance between the start point and
end point, no matter what route was traveled between them.

In this case, the displacement is 15 meters east.
So the velocity is

(15 meters east) / (5 sec) = 3 meters/sec east .

Answer:

i dont know what he said but i think C

Explanation:

I need help I can't solve it and I'm struggling badly. Justin weighs 185 lb. and decides to climb a staircase that has a vertical distance of 12 m. He climbs the stairs in 12.7 seconds. Calculate his work:

Answers

Answer:

Power is work/time - not needed here

W(work) = F * D force * distance

Here W = M g h

Since 2.20 lb is equivalent to 1 kilogram mass

W = 185 lb / 2.20 lb/kg * 9.80 m/s^2 * 12 m = 824 Joules

note that a pound is a unit of force and a kilogram is a unit of mass

Organisms that live in desert and desert-scrub biomes have developed unique adaptations that aid in their survival. The Sahara Desert takes up ten percent of the continent of Africa, and is the largest desert in the world. The species that live in the Sahara are highly adapted to survive in the extreme conditions they live in. What morphological and physiological adaptations might be seen in the organisms that live in the Sahara Desert?

Answers

Some morphological and physiological adaptations of organisms includes special storage organs of geophytes and other plants, deep root systems of trees in order to reach deep aquifers. Additionally, some plants survive by germinating quickly after the rains and before the summer heat these plants complete their life cycles.

Answer:

C. small bodies and long limbs in warm-blooded organisms, light colored fur or feathers to reflect sunlight, lack of sweat glands in many organisms, hibernation during hot months

Explanation:

Tekanan gas yang berada dalam suatu ruangan dikurangi hingga setengahnya.hal ini menyebabkan suhu gas tersebut turun menjadi 3/4 dari semula.tentukanlah perbandingan volume gas akhir dan volume gas awal

Answers

PV/T =C

P.V1/T = 1/2P.V2/3/4T
V1= 4V2/6
V2/V1 = 6/4 =3/2

if the earth stops rotating about its axis then what will be the change in the value of g at a place in the equator-plane? Assume that earth is a sphere with a radius of 6.4*10*6m.

Answers

Answer:

if the earth stops rotating the value of g increases

Phileas Fogg, the character who went around the world in 80 days, was very fussy about his bathwater temperature. It had to be exactly 38.0o C. You are his butler, and one morning while checking his bath temperature, you notice that it’s 42.0oC. You plan to cool the 100.0 kg of water to the desired temperature by adding an aluminum-duckie originally at freezer temperature (-24.0oC). Of what mass should the Al-duckie be? [Specific heat of Al = 0.900 J/(goC); density of water =1 .00 g/ml]. Assume that no heat is lost to the air

Answers

Answer:

The mass of the Al-duckie should be 30 kg.

Explanation:

We will use the first law of thermodynamics:

ΔU = m·Cv·ΔT

Since the specific heat of water is 4.185 J(gºC), the change in the water's internal energy would be:

ΔU = 100 kg · 4.185 J(gºC) · (42ºC - 38ºC) = 1674 KJ

Given that no heat is lost, all the internal energy that the water loses while cooling down will transfer to the duckie. So, if the duckie has ΔU = 1674 KJ and its final temperature is the desired 38 ºC, we can calculate its mass using the first law again:

$m=\frac{\Delta{U}}{Cv{\Delta{T}}}=(1674)/(0.9*[38-(-24)])=30Kg$

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