All categories

3 years ago

I need help with these questions.

Physics

1 answer:

strojnjashka [21]3 years ago

4 0

I'll go ahead and answer the ones here without an answer. For reference, the half-life formula is <em>final amount = original amount(1/2)^(time/half-life)</em>

<em />

4) 12.5g

x = 100(1/2)^(63/21)

5) 50g

3.125 = x(1/2)^(0.1/0.025)

6) 500g

x = 4000(1/2)^(525/175)

7) 0.24g

0.06 = x(1/2)^(11430/5730)

8) 125g

x = 1000(1/2)^(17100/5700)

Hope this helps! :)

You might be interested in

You drop a rock off the top of a 100 m tall building. How long does it take to hit the ground?

Vitek1552 [10]

It would mostly depend on its weight

4 0

3 years ago

Which of these best describes the difference between energy and power? A) Power and energy have nothing in common B) Energy and

Sphinxa [80]

D.Power has a time component while energy does not. This is because power is the RATE at which work is performed.

8 0

3 years ago

A steel ball bearing is released from a height H and

ArbitrLikvidat [17]

Answer:

ELASTIC collision

kinetic energy is conservate

Explanation:

As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.

As the floor does not move the conservation of the moment

po = pf

-mv1 = m v2

- v1 = v2

So the speed with which it descends is equal to the speed with which it rises

Therefore the kinetic energy of the ball before and after the collision is the same

4 0

3 years ago

1. Why is it important to use units in any graph?

Scorpion4ik [409]

Answer:

so its easier to understand for the reader

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll

Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

$Q = \dfrac{4mME }{(m+M)^2}$

However; from the total stopping power & power loss of the electron;

$\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}$

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss $=\dfrac{82 \times 1.9}{800}$

= 0.19475

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

4 0

3 years ago