3 years ago

How long must you wait (in half-lives) for a radioactive sample to drop to 2.10 % of its original activity?

Physics

1 answer:

MArishka [77]3 years ago

6 0

Answer:

222/88 Ra

Explanation:

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Describe the interaction with the sun and producers

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When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to

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Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

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d is the separation between the two charges

We can also rewrite it as

$d=\sqrt{k\frac{q_1 q_2}{F}}$

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

$d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}$

so the correct option is E.

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3 years ago

In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls

sweet-ann [11.9K]

Answer:

Acceleration= $0.5m/s^2$

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration= $a=\frac{F}{m}$

Using the formula

Acceleration of box= $\frac{20}{40}=0.5m/s^2$

The acceleration of the box= $0.5m/s^2$

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

$a=\frac{30}{40}=\frac{3}{4} ms^{-2}$

$v^2-u^2=2as$

Substitute the values

$v^2-0=2\times \frac{3}{4}\times 0.3=0.45$

$v^2=0.45$

$v=\sqrt{0.45}=0.67m/s$

Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s

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3 years ago