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3 years ago

What are the difference between conductor and insulator?

Physics

1 answer:

Anvisha [2.4K]3 years ago

5 0

Answer:

A conductor allows energy e.g. electric charge or heat to pass through it easily. While an insulator does not let electric current or heat to travel through it.
Insulators have strong molecular bonds. While molecular bonds are very weak in conductors.
Insulators have very low conductivity. While in conductors, it is very high.
Insulators have a very high resistance and therefore the electrons are held together very firmly. The conductors, on the other hand, have a very low resistance.
Insulators do not have any electric field, neither inside nor on the surface. While in conductors, it is found on the surface and continues to be zero in the inner part of the conductor.

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If a liter of water is heated from 20 C to 50 C, what happens to its volume?

iragen [17]

Answer:

The volume increases

Choice B is correct

Explanation:

If a liter of water is heated from 20 C to 50 C its volume will increase.

Matter exists in 3 states;

Solids

Liquids

Gases

Matter expands when heated, consequently occupying more space. This implies that volume will increase

5 0

3 years ago

Match each term to its definition.

ladessa [460]

Ion is e
Element is a
Compound is f
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4 years ago

Use the table below to answer the following questions. Substance Specific Heat (J/g•°C) water 4.179 aluminum 0.900 copper 0.385

lbvjy [14]

1. $-8.78 \cdot 10^5 J$

The energy lost by the water is given by:

$Q=m C_s \Delta T$

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

$\Delta T=10.0C-80.0C=-70.0 C$ is the change in temperature

Substituting,

$Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J$

2. $3.24 \cdot 10^4 J$

The energy added to the aluminium is given by:

$Q=m C_s \Delta T$

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

$\Delta T=150.0 C-30.0C =120.0 C$ is the change in temperature

Substituting,

$Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J$

3a. $-5.6^{\circ}C$

The temperature change of the water is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = -232 kJ=-2.32\cdot 10^5 J$ is the heat lost by the water

$m=10.0 kg=10000 g$ is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C$

3b. $+10.2^{\circ}C$

The temperature change of the copper is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = 1.96 kJ=1960$ is the heat added to the copper

$m= 500 g$ is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C$

4. 42.9 g

The mass of the water sample is given by

$m=\frac{Q}{C_S \Delta T}$

where

$Q=4300 J$ is the heat added

$\Delta T=39 C-15 C=24C$ is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g$

5. 115.5 J

The heat used to heat the copper is given by:

$Q=m C_s \Delta T$

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

$\Delta T=80.0 C-20.0C =60.0 C$ is the change in temperature

Substituting,

$Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J$

6. 0.185 J/g•°C

The specific heat of iron is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

$\Delta T=25.0-50.4 C=-25.4 C$ is the change in temperature

Substituting,

$C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC$

8 0

4 years ago

Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

Learn more about accelerated motion:

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4 years ago

A block of mass m rests on a frictionless, horizontal surface. A cord attached to the block passes over a pulley whose radius is

ra1l [238]

Answer:

Explanation:

Let the tension in the cord be T₁ and T₂ .

for motion of block placed on horizontal table

T₁ = m a , a is acceleration of the whole system .

for motion of hanging bucket of mass m

mg - T₂ = ma

adding the two equation

mg + T₁- T₂ = 2ma

for rotational motion of the pulley

torque = moment of inertia x angular acceleration

(T₂ - T₁) R = I x α , I is moment of inertia of pulley , α is angular acceleration .

(mg - 2ma ) R = I x α

(mg - 2ma ) R = I x a / R

(mg - 2ma ) R² = I x a

mgR² = 2ma R² + I x a

a = mgR² / (2m R² + I )

Since body moves by distance d in time T

d = 1/2 a T²

a = 2d / T²

mgR² / (2m R² + I ) = 2d / T²

mgR²T² = 2d x (2m R² + I )

mgR²T² - 4dm R² = 2dI

m R² ( gT² - 4d ) = 2dI

I = m R² ( gT² - 4d ) ] / 2d .

3 0

3 years ago

What are the difference between conductor and insulator? ​

What are the difference between conductor and insulator?