Ask question

Ask question

All categories

3 years ago

14

A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material suc

h as rubber is a(n) _______________ because it _______________ the flow of electric charges. A material that partially conducts electric current is a(n) _______________. These materials include _______________ elements.

2 answers:

notka56 [123]3 years ago

6 0

Explanation:

A metal such as copper is a <u>conductor</u> because it provides a pathway for electric charges to move easily. A material such as rubber is an <u>insulator</u> because it <u>resists</u> the flow of electric charges. A material that partially conducts electric current is a <u>semiconductor</u>. These materials include <u>group 3 and group 5</u> elements.

jeka57 [31]3 years ago

4 0

Answer:

conductor

insulator

resists

semiconductor

group 3 and group 5

Explanation:

You might be interested in

As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th

Gre4nikov [31]

Transvere wave because the direction which the particles are being displaced

5 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

where:

$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

$v_2=0$ is the final velocity (the car comes to a stop)

$t_2=5 s$ is the time elapsed in the 2nd phase

Substituting,

$s_2=\frac{32.4+0}{2}(5)=81 m$

So, the total displacement of the car is

$s=s_1+s_2=324+81=405 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago

The distance around the block is equal to 3000 feet, how many blocks would you have to run to run a total of a mile?

lesya692 [45]

You would have to run a little less than 2 blocks

7 0

3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater

Mice21 [21]

Answer:

So, at the depth of 24 cm below the surface of the glycerine the pressure is 2970 Pa. Hence, this is the required solution.

Explanation:

Given that,

Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.

The density of glycerine,

We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :

h = 0.24 meters

or

h = 24 cm

5 0

3 years ago