Ask question

Ask question

All categories

Hunter-Best [27]

3 years ago

14

Testing shows that a sample of wood from an artifact contains 12.5% of the original amount of carbon-14. Given that the half-lif

e of carbon-14 is 5730 years, how old is the artifact?
A. 22,920 years
B. 17,190 years
C. 11,460 years
D. 5730 years

2 answers:

castortr0y [4]3 years ago

7 0

Answer: i think

Explanation:

d

ratelena [41]3 years ago

5 0

Answer:

B. 17,190 (A P E X)

Explanation:

If the half-life of carbon-14 is 5730 years then for every 5730 years the original amount of the carbon-14 is cut in half. So if it originally started at 100% when it was new then in 5730 years it would drop to 50%. After that, every 5730 years it drops to 25% and then 12.5% and so on.

You might be interested in

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago

A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

<u>Where:</u>

<em>N = is the number of turns = 1 </em>

<em>ΔΦ = ΔB*A </em>

<em>Δt = is the time = 0.3 s </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T </em>

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

4 years ago

Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto

timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

Vector B = ?

Resultant of vector A and B = R = -14j

R = A + B

-14j = 29j + B

B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

4 0

3 years ago

What is the mass of a ball that has 29j of potential energy and is lifted 2.0m?

Salsk061 [2.6K]

Answer:

1.48kg

Explanation:

Here,

potential energy (P.E) = 29j

height (h) = 2m

acceleration due to gravity(g) =

$9.8m {s}^{ - 2}$

mass(m) = ?

we know,

P.E = mgh

or, 29 = m×9.8×2

or, 29/19.6 = m

or,m = 1.48kg

6 0

2 years ago

A collapsible plastic bag (figure below) contains a glucose solution. If the average gauge pressure in the vein is 1.29 A collap

expeople1 [14]

Answer:

0.125 m

Explanation:

Pressure in fluids is given as the product of density, height and acceleration due to gravity and expressed as

P=hdg

Where h is the height, d is density, g is acceleration due to gravity and P is pressure.

Making h the subject of formula then

h=P/dg

Given specific gravity of a substance, its density is equal to specific gravity multiplied by density of water. Taking density of pure water as 1000 kg/m³ then the density of reference fluid will be 1.05*1000=1050 kg/m³

Substituting pressure with 1.29*10³ pa as given then taking g as 9.81 m/s² then

H=1.29*10³÷(9.81*1050)=0.1252366389981068880151448958788408329692m

Rounded off, the height is approximately 0.125 m

8 0

4 years ago