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3 years ago

14

Carbon-14 has a half life of 5730 years. Consider a sample of fossilized wood that when alive would have contained 24g of C-14 i

t now contains 1.5g. How old is the sample

1 answer:

kirza4 [7]3 years ago

6 0

Answer:

5732 years

Explanation:

hope it helps, if not im sorry

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DO not answer unless you actually now how to do this!!

Digiron [165]

Ep max = 6.825 J
Change of Ep = 4.425 J

3 0

3 years ago

a solution made by dissolving 116 g of CaCl2 in 64 g of water has a density of 1.180 g/ml at 20 degrees celsius what is the mola

Makovka662 [10]

Answer:

The molarity of CaCl2 in the solution is 4.94 M

Explanation:

First of all you need to calculate CaCl2 mass.

You have one atom of Ca = 40.07 g/mol and two atoms of Cl = 35.45 g/mol so the molecule has a mass of 110.97 g/mol.

Now, knowing that your solution will have a mass of 64 grams of water + 116 grams of CaCl2 = 180 grams, you can calculate its volume, knowing that density = mass/volume

density x mass = volume --> 1,180 g/ml x 180 g = 212.4 ml

In 212.4 ml, you have 116 grams of CaCl2. You can calculate how many moles of CaCl2 you have:

110.97 g ------ 1 mol

116 g -------- x = (116 g x 1 mol) / 110.97 g = 1.05 moles

The molarity in a solution equals how many moles of a certain solute you have in 1000 ml of solution. In this solution, you have 1.05 moles in 212.4 ml, so in 1000 ml you will have:

212.4 ml ------- 1.05 moles

1000 ml -------- x = (1000 ml x 1.05 moles) / 212.4 ml = 4.94 moles.

This means the molarity of CaCl2 in the solution is 4.94 M.

4 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

<u>r < a:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

<u>r = a:</u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>a < r < b:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>r = b:</u>

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

<u>r < c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u /> $E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$ <u />

<u>At r = b:</u>

<u /> $E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$ <u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>At r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

What is your initial speed if you accelerate at 5.8 m/s/s for 3.0 seconds and achieve a final speed of 45 m/s?

Natali5045456 [20]

Answer:

27.6 m/s

Explanation:

hopefully it makes sense and is visible

:)

8 0

2 years ago

what occurs when a swimmer pushes through the water to swim answers are (A) the water exerts a reaction force on the swimmer (B)

miv72 [106K]

When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.

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3 years ago

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