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2 years ago

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carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr

oximately how many years ago was this shell part of a living oranism?

1 answer:

denis23 [38]2 years ago

3 0

The half-life equation $m=m_{0} (\frac{1}{2})^n$ in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>

<em> $n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}$ </em>

<em> $\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2$ </em>

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

<em>2 * 5700 = </em>11400 yr

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An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is held in place, how fast will the elec

inna [77]

Answer:

v = 61.09m/s

Explanation:

In order to calculate the speed of the electron when it is 3.00cm from the proton, you first calculate the acceleration of the electron, produced by the electric force between the electron and the proton. By using the second Newton law you have:

$F=ma=k\frac{q^2}{r^2}$ (1)

m: mass of the electron = 9.1*10^-31kg

q: charge of electron and proton = 1.6*10^-19C

r: distance between electron and proton = 9.00cm = 0.09m

k: Coulomb's constant = 8.98*10^9Nm2/C^2

You solve the equation (1) for a, and replace the values of the other parameters:

$a=\frac{kq^2}{mr^2}=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)^2}{(9.1*10^{-31}kg)(0.09m)^2}=3.11*10^4\frac{m}{s^2}$

Next, you use the following formula to calculate the final speed of the electron:

$v^2=v_o^2+2ax$ (2)

vo: initial speed of the electron = 0m/s

a: acceleration = 3.11*10^4m/s^2

x: distance traveled by the electron

When the electron is at 3.00cm from the proton the electron has traveled a distance of 9.00cm - 3.00cm = 6.00cm = 0.06m = x

You replace the values of the parameters in the equation (2):

$v=\sqrt{2ax}=\sqrt{2(3.11*10^4m/s)(0.06m)}=61.09\frac{m}{s}$

The speed of the electron is 61.09m/s

8 0

2 years ago

You hang an object with mass m = 0.380 kg from a vertical spring that has negligible mass and force constant k = 60.0 N/m. You p

joja [24]

Answer:

a) 0.500 s

b) greater than 0.500 s

c) greater than 0.500 s

Explanation:

The time period of an oscillating spring-mass system is given by:

$T=2\pi \sqrt{\frac{m}{k}}$

where, m is the mass and k is the spring constant.

a) As the period of oscillation does not depend on the distance by which the mass is pulled, the period would remain same as 0.500 s for the given system.

b) As the period varies inversely with the square root of spring constant, so with the decrease in the spring constant, the period would increase. So, the new period would be greater than 0.500 s.

c) As the period varies directly with the square root of mass, so with the increase in mass, the period will also increase. The new period will be greater than 0.500 s.

8 0

2 years ago

Mr. Dunn drives 64.8km from work at a speed of 48km/h. Mrs. Dunn drives 81.2km from work

artcher [175]

Answer:

i) Mr. Dunn arrives to home first.

ii) 3 min

Explanation:

i. To find who arrives first to home you calculate the time, by using the following formula:

$t=\frac{x}{v}$

x: distance

v: velocity

Mr. Dunn:

$t=\frac{64.8km}{48km/h}=1.35h$

Mrs. Dunn:

$t=\frac{81.2km}{58km/h}=1.4h$

Hence, Mr. Dunn arrives to home first.

ii. To calculate the difference in minutes, you convert hours to minutes:

$1.35h*\frac{60min}{1h}=81min\\\\1.40h*\frac{60min}{1h}=84min\\\\\Delta\ t=(84-81)min=3min$

the difference between the times is 3min

7 0

2 years ago

Hey guys I really really need help with this question for ASAP! Explain what chart junk is and how it differs from the kind of i

Ahat [919]

I found a definition: Chartjunk<span> refers to all visual elements in </span>charts<span> and graphs that are not necessary to comprehend the information represented on the graph, or that distract the viewer from this information.</span>

3 0

2 years ago

Read 2 more answers

A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di

Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

$v_f = 12 m/s$

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

$\int (80 - (40/3)t) dt = 80(v_f - 12)$

$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$

6 0

2 years ago