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antoniya [11.8K]
3 years ago
12

If we start with 400 atoms of a radioactive substance, how many would remain after one half-life? After 2 half-lives? After 3 ha

lf-lives? After 4 half-lives?
Physics
1 answer:
AysviL [449]3 years ago
3 0

Answer:

how many would remain

after one half-life = 200 atoms

After 2 half-lives = 100 atoms

After 3 half-lives = 50 atoms

After 4 half-lives = 25 atoms

Explanation:

Given;

Initial amount = 400 atoms

Half life is the time taken to decay half of a radioactive material.

how many would remain after one half-life;

= 400/2 = 200 atoms

After 2 half-lives;

= 400/2^2 = 400/4

= 100 atoms

After 3 half-lives

= 400/2^3 = 400/8

= 50 atoms

After 4 half-lives

= 400/2^4 = 400/16

= 25 atoms

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A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
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4 0
4 years ago
A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas
Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

3 0
3 years ago
Which of the following is the best description of a gaseous substance?
sveta [45]

Answer:

It has no shape of its own but has a definite volume.

Explanation:

Gases have no shape but a definite volume

6 0
3 years ago
Read 2 more answers
Please help me with this please <br><br> I’ll mark you Brainly
ivolga24 [154]
I think it is option (C).

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4 0
3 years ago
A 2.31 kg rope is stretched between supports 10.4 m apart. If one end of the rope is tweaked, how long will it take for the resu
zlopas [31]

Answer:

t = 0.657 s

Explanation:

First, let's use the appropiate equations to solve this:

V = √T/u

This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.

Where:

V: Speed of the disturbance

T: Tension of the rope

u: linear density of the rope.

The density of the rope can be calculated using the following expression:

u = M/L

Where:

M: mass of the rope

L: Length of the rope.

We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:

u = 2.31 / 10.4 = 0.222 kg/m

Now, replacing in the first equation:

V = √55.7/0.222 = √250.9

V = 15.84 m/s

Finally the time can be calculated with the following expression:

V = L/t ----> t = L/V

Replacing:

t = 10.4 / 15.84

t = 0.657 s

4 0
3 years ago
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