All categories

4 years ago

A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?

Physics

1 answer:

lesantik [10]4 years ago

3 0

<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>

Half life is the time taken for a radioactive isotope to decay by half of its original amount.
We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

You might be interested in

Which of the following is equal to Velocity change per second

Mademuasel [1]

Acceleration

Explanation:

Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.

Acceleration = $\frac{V - U }{T}$

V is the initial velocity

U is the final velocity

T is the time

It is has a unit of m/s²

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

8 0

3 years ago

The energy transfer diagram represents the energy of a person diving into a pool.

GREYUIT [131]

The answer is A) 1000 J
According to the law of conservation of energy, the total energy before an action is always equal to the total energy after the action.
So that is, the total energy is 8000J found as potential energy, 7000J has transformed into kinetic energy, then the thermal energy should be the remaining 1000J.

Hope this helps.

6 0

3 years ago

Which parts of the graph would increase if all of earth's polar ice melts? Which part would decrease?

Sedbober [7]

Answer:

the answer to the question is the part that is reflected by clouds and atmosphere 26%

8 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

$\frac{x}{D}=\frac{1.22\lambda }{d}$