All categories

3 years ago

Jason makes a graphic organizer for insulators and conductors and wants to add the following labels to it.

Physics

2 answers:

FrozenT [24]3 years ago

7 0

Jason should place the regions in:

C) Label 1: X

Label 2: Z

MA_775_DIABLO [31]3 years ago

4 0

Answer: C) Label 1: X

Label 2: Z

Explanation: Conductors are those substances which conduct electrical current easily because of the presence of free electrons whereas Insulators do not conduct electrical current as they do not contain free electrons. Some common examples of conductors are copper, aluminum, gold, and silver. Some common examples of insulators are glass, air, plastic, rubber, and wood.

Label 1: Lacks free electrons must be placed in the area marked X as they correspond to insulators.

Label 2: Aluminum is an example of conductor and thus must be labelled in area marked by Z.

Area Y corresponds to semiconductors which have properties intermediate to conductors and insulators.

You might be interested in

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

$v^2-u^2 = 2gd$

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

$v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s$

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

$y_1 = h - \frac{1}{2}gt^2$

While the vertical position of the ball thrown upward is

$y_2 = ut - \frac{1}{2}gt^2$

The two balls meet when

$y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s$

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

$y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m$

So the distance below the top of the cliff is

$d=24.0 - 18.0 = 6.0 m$

4 0

3 years ago

Which two types of information are found in an elements box in the periodic table

Pani-rosa [81]

Answer:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name.

Explanation:

4 0

3 years ago

Read 2 more answers

The rate at which temperature decreases with height in the atmosphere is called the

Genrish500 [490]

It is called <span>environmental lapse rate</span>

5 0

3 years ago

May I please get some help?

xxMikexx [17]

B is the correct q2ureire awej\

3 0

3 years ago

An 18g bullet is shot vertically into a 10kg block. The block lifts upward 9mm. The bullet penetrates the block in a time interv

stealth61 [152]

Answer:

The initial kinetic energy of the bullet is closest to 491.87 J

Explanation:

Given;

mass of bullet, m₁ = 18g = 0.018kg

mass of block, m₂ = 10kg

height moved by the block, h = 9 mm = 0.009 m

time taken for the bullet to travel through the block, t = 0.001s

let the initial velocity of the bullet = v₁

let the final velocity of the bullet = v₂

Apply the principle of conservation of linear momentum;

initial momentum = final momentum

0.018v₁ = v₂(0.018 + 10)

0.018v₁ = 10.018v₂ -----equation (1)

Apply the law of conservation of energy when the bullet lifts the block through 9mm

mgh = ¹/₂mv₂²

gh = ¹/₂v₂²

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.009)

v₂ = 0.42 m/s

Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;

0.018v₁ = 10.018v₂

0.018v₁ = 10.018(0.42)

0.018v₁ = 4.208

v₁ = 4.208 / 0.018

v₁ = 233.78 m/s

Now, determine the initial kinetic energy of the bullet;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(0.018)(233.78)²

K.E₁ = 491.87 J

Therefore, the initial kinetic energy of the bullet is closest to 491.87 J

6 0

3 years ago