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2 years ago

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A long cylindrical insulator has a uniform charge density of 1.54 µc/m3 and a radius of 6cm what is the electric field inside th

e insulator at a distance of 2 cm

1 answer:

Lubov Fominskaja [6]2 years ago

4 0

Plug given values in SI units to get electric field.

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Using the periodic table, choose the more reactive metal. Al or Mg

ziro4ka [17]

Aluminum, and magnesium are metals. For metals, reactivity decreases as you go from left to right across the periodic table. Atomic number of Al is 13 and of Mg is 12. Hence the least reactive of these two is therefore aluminum.

Magnesium is "HIGHLY FLAMMABLE" carefully take a small piece and hit it with a torch. If its Magnesium it will "Caution, very, quickly burn.
Aluminum will not react to simple flame, it will only melt with enough direct heat.

Magnesium
==========
Atomic Number: 12
Atomic Symbol: Mg
Atomic Weight: 24.305
Electron Configuration: 2-8-2

Aluminum
========
Atomic Number: 13
Atomic Symbol: Al
Atomic Weight: 26.9815
Electron Configuration: 2-8-3

Hope this helps some. Any questions please feel free to ask. Thank you

6 0

2 years ago

Read 2 more answers

Our city is considering a program to spray insecticide to reduce the number of mosquitoes in the area. Before they do, they want

Anarel [89]

(C) the government agency that regulates these types of chemicals

3 0

2 years ago

Read 2 more answers

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

2 years ago

What are the unit for acceleration

Flura [38]

<h3>Answer</h3>

m/s^2 (meter per sec square)

Explanation:

acc = change in velocity/time

= distance/time

----------------

time

= m/s

------

s

=m/s^2

7 0

2 years ago

A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th

Pavel [41]

Answer:

The block's mass should be $3m$

Explanation:

Given:

Cart with mass $m$

From the conservation of energy before mass is added,

$\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}$

Where $A =$ amplitude of spring mass system, $k =$ spring constant

$A = v\sqrt{\frac{m}{k} }$

Now new mass $M$ is added to the system,

$\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}$

$A = v \sqrt{\frac{m +M }{k} }$

Here, given in question frequency is reduced to half so we can write,

$f' = \frac{f}{2}$

Where $f =$ frequency of system before mass is added, $f' =$ frequency of system after mass is added.

$\omega ' = \frac{\omega}{2}$

$\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}$

$\frac{k}{m +M } = \frac{k}{4m}$

$M = 3m$

Therefore, the block's mass should be $3m$

8 0

2 years ago