The buret clamp was used to stabilize the

A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field

maksim [4K]

Answer:

the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ )

Substituting the values into the equation, it becomes

E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ ) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²( $1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }$ ) = 7.12 × 10⁵( $1 - \frac{0.12}{0.1216}$ ) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

7 0

3 years ago

When CO2 and H2O react to form H2CO3, the number of bonds broken in H2O is ____ and in CO2 is _____. one; one

guapka [62]

CO2 and H2O react to form H2CO3 and two bonds are broken each in CO and H2O to form H2CO3.

<h3>What is chemical bonding?</h3>

Chemical bonding refers to the forces of attraction which hold atoms of the same or different elements together in order to form stable compounds or molecules .

Chemical bonding may be either ionic or covalent.

The greater the number of bonds in a compound, the more stable the compound.

During chemical reactions, bonds are broken and new binds are formed.

There are two bonds each in CO2 and H2O.

This, in the reaction between CO2 and H2O react to form H2CO3, , the number of bonds broken in H2O is two and in CO2 is two.

Learn more about chemical bonding at: brainly.com/question/819068

5 0

2 years ago

Is MG an ion, isotope, or neutral atom?

Kitty [74]

Answer:

isotope

5 0

3 years ago

Based on the diagram below, how much of the excess reactant is left over? *

pantera1 [17]

Answer:

3 pieces of lunch-meat and 2 slices of cheese

Explanation:

You have enough bread to make 3 sandwiches

You have enough lunch-meat to make 4 sandwiches

You have enough cheese to make 5 sandwiches

In all you have enough material to make 3 sandwiches

so if you subtract three from each number above you will have no bread, enough lunch-meat to make one sandwich and enough cheese to make two sandwiches

luch-meat for one sandwich is: 3 pieces

Cheese for two sandwiches is: 2 pieces

5 0

3 years ago

Could someone please care to explain what the answer is

polet [3.4K]

Calculate the percentage composition of Ca(MnO4)2.

<u>Firstly, we see how many atoms are there for each element in the formula</u>.

Ca= 1 atom

Mn= 2 atoms

O= 8 atoms

<u>Next, we are going to consult our periodic table for the atomic mass of each element.</u>

Ca= 40

Mn= 55

O= 16

Then, we have to find the molar mass for the compound..

Here is the formula for calculating molar mass of an element:

Molar Mass= ( no. of atoms of the element × atomic mass of the element)

Now, we have to calculate the atomic mass of the compound. So using the molar mass formula for an element, we calculate the molar mass for each element then we sum up their molar masses to get the compounds molar mass.

Molar mass (Ca)= 1× 40

(Ca)= 40

Molar mass (Mn)=2×55

(Mn)= 110

Molar mass (O)= 8×16

(O)= 128

Now: Molar mass( compound)= (Ca)+(Mn)+(O)

= 40+ 110 128

= 278

This is everything we need to calculate our percentage composition for each element..

* The example says to find the percentage composition for Ca. So we only find for Ca, Which is already done using the formula and the answer is 14.39%.

To prove that your answer is correct, find the percentage composition for Mn and O as well. Then you add up their percentage compositions.

If you do and you get 100 as your answer, then your percentage compositions are correct.

Why don't you try finding the percentage composition for Mn and O, then add up all the three percentage compositions. If you 100 as their sum, then your percentage composition for each of the elements are correct.

6 0

3 years ago