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2 years ago

Why do conductors transfer heat faster than insulators?

Physics

2 answers:

ryzh [129]2 years ago

7 0

Because they have freely moving electrons that can transfer thermal energy quickly and easily

Nesterboy [21]2 years ago

3 0

Answer:

Conduction is usually faster in certain solids and liquids than in gases. Materials that are good conductors of thermal energy are called thermal conductors. Metals are especially good thermal conductors because they have freely moving electrons that can transfer thermal energy quickly and easily.

Heat transfer by convection happens through the air, and there are millions of minuscule air spaces between the fibers. Heat transfer by radiation is also slow since one fiber must radiate its heat to another.

When we give heat then kinetic energy is increase and this heat is transferred from hot metal to cold metal through this free electrons. As in insulator the free electrons are negligible so that the heat is not transferred from hot junction to cold junction due to absence of this free electrons.

Explanation:

maek me as brainliest

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1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.

Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

$v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h$

Hence, this is the required solution.

8 0

4 years ago

A skateboarder initially has 5 kJ of kinetic energy. As she freewheels along a flat section of path, she does 400 J of work agai

sweet [91]

The final kinetic energy of the skateboarder after she freewheels and did work against friction on the flat section of the path is 4,600 J.

<h3>Conservation of energy</h3>

The final kinetic energy of the stakeboarder is determined by applying the principle of conservation of energy as shown below;

ΔK.E = -W

K.Ef - K.Ei = -W

where;

K.Ef is the final kinetic energy
K.Ei is the initial kinetic energy
W is work done

K.Ef = K.Ei - W

K.Ef = 5,000 J - 400 J

K.Ef = 4,600 J

Thus, the final kinetic energy of the skateboarder is 4,600 J.

Learn more about kinetic energy here: brainly.com/question/25959744

4 0

2 years ago

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago

A skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. How much higher is he when his velocity has red

Taya2010 [7]

Mechanical energy E = mgh + 1/2mv²

When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²

Without friction, energy is conserved at all times.

E₁ = E₂
↓
1/2mv₁² = mgh + 1/2mv₂²
↓
1/2v₁² = gh + 1/2v₂²
↓
gh = 1/2(v₁² - v₂²)
↓
h = (v₁² - v₂²) / (2g)

5 0

4 years ago

Read 2 more answers

acontainer is filled whith mercury to alevel of 10m whit water to alevel of 8m and whit oil to alevel of 5m the densities oil ,w

Alborosie

Answer:

1450.4 KN $m^{2}$

Explanation:

Pressure = ρhg

where: ρ is the density of the liquid, h is the height and g the force of gravity.

Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury

So that,

i. Pressure of oil = ρhg

(ρ = 0.8 g/cm³ = 800 kg/m³)

= 800 x 5 x 9.8

= 39200

Pressure of oil = 39200 N $m^{2}$

ii. Pressure of water = ρhg

(ρ = 1 g/cm³ = 1000 kg/m³)

= 1000 x 8 x 9.8

= 78400

Pressure of water = 78400 N $m^{2}$

ii. Pressure of mercury = ρhg

(ρ = 13.6 g/cm³ = 13600 kg/m³)

= 13600 x 10 x 9.8

= 1332800

Pressure of mercury = 1332800 N $m^{2}$

So that,

Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800

= 1450400

= 1450.4 KN $m^{2}$

Total pressure exerted by the liquids at the base is 1450.4 KN $m^{2}$ .

8 0

3 years ago