___increases until a___

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that

almond37 [142]

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

Half-Life refers to the time it takes for half the amount of a substance to disappear or change.

The nucleus of the atoms of radioactive elements disintegrate to half their starting amounts after every Half-Life.

After three half-lives one-eight of the original atoms remain.

Therefore, after three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

Learn more about Half-Life at: brainly.com/question/26689704

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7 0

2 years ago

The reason carbon dating works is that A. there is so much non-radioactive carbon dioxide in the air. B. when a plant or animal

yuradex [85]

D. There is a known constant concentration of C14 in Nature. As we consume living things to survive our bodies (made of carbon) stop replenishing our body's carbon (we stop eating) and start to decay. Since we know the 1/2 life of C14 and the ratio of C14 to normal C12 we can determine fairly accurately how long ago a thing stopped consuming carbon (e.g. when it died)

7 0

3 years ago

Alice and Bob sit on a see saw, that consists of a uniform beam of mass 20kg and a fulcrum rock, that acts as a pivot point. The

tankabanditka [31]

Answer:

Explanation:

Given

Mass of alice $M_a=50 kg$

Mass of bob $M_b=70 kg$

Mass of beam $M=20 kg$

Suppose fulcrum is at distance of d from bob thus From diagram

Equating torque i.e. torque of Bob,Alice and beam must cancel out each other to be balanced

$500\times (8-d)+20\times (4-d)=70\times d$

$400-50d+80-2d=70d$

$480=140d$

$d=\frac{480}{140}$

$d=3.43 m$

5 0

4 years ago

Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>please

REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

$6=2+a*2\\4=2*a\\a=2[m/s^{2} ]$

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

$25=5+2*t\\t = 10 [s]$

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

$v_{f}=4+10*2\\v_{f}=24 [m/s]$

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$v_{f}=5+8*10\\v_{f}=85 [m/s]$

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0$

8 0

3 years ago

Ring of mass straight m and radius straight r is attached to the end of a thin rod of mass straight m and radius 2 straight r

Katen [24]

The total moment of inertia about an axis is : $L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}$ for a ring of mass m and radius straight r attached to a thin rod.

<h3>Determine the Total moment of Inertia about an axis </h3>

<u>Given data:</u>

mass of ring --> m

radius of ring --> r

mass of rod --> M

Length of rod ---> L ( 2 * radius )

Total Moment of Inertia about an axis = Irod + Iring

where : Irod = moment of inertia of rod, Iring = moment of inertia of ring

Irod = ML² / 3

Iring = 2mr² / 5

moment of inertia around an axis by Iring = I

where ; I = 2mr² / 5 + ML² according to parallel axis theorem

Hence the Total moment of Inertia about an axis is :

Itotal = 2mr²/5 + ML² + ML² / 3

= $L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}$

Learn more about Moment of inertia : brainly.com/question/6956628

4 0

3 years ago

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