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Vika [28.1K]
4 years ago
11

Examine the false statement. Scientists can directly observe the half-life of all radioactive elements. What option rewords the

statement so that it is true? Select all that apply. Scientists can calculate the rate of nuclear decay for all radioactive elements using half-life. Scientists can calculate the half-life of all radioactive elements using the rate of nuclear decay. Scientists cannot directly observe or calculate the rate of nuclear decay for all radioactive elements. Scientists can directly observe the rate of nuclear decay for all radioactive elements.
Physics
2 answers:
Bingel [31]4 years ago
7 0

Answer:

Scientists can directly observe the rate of nuclear decay for all radioactive elements.

Scientists can calculate the half-life of all radioactive elements using the rate of nuclear decay

Explanation:

I just took a test and got it right

Ivenika [448]4 years ago
6 0

Scientists can directly observe the rate of nuclear decay for all radioactive elements.

Scientists can calculate the half-life of all radioactive elements using the rate of nuclear decay

Explanation:

The true statements from the given options is that "scientists can directly observe the rate of nuclear decay for all radioactive elements" and "scientists can calculate the half-life of all radioactive elements using the rate of nuclear decay".

Let us carefully examine the original statement:

Scientists can directly observe the half-life of all radioactive elements

The keywords from here are:

       directly observe

        half-life

       radioactive elements

to directly observe implies that evaluative information about the half-life is being gathered. In making observations, calculations can be made to make deductions.

half-life is the time taken for half of radioactive nuclei to disintegrate.

Radioactive elements  are unstable nuclei that will naturally decay on their own into other nuclei.

It is correct to say that scientist can directly observe the rate of nuclear decay for all radioactive element. The half-life is one of the ways of measuring the rate of decay.

Scientists can calculate the half-life of all radioactive elements using the rate of nuclear decay also.

Learn more:

Half-life brainly.com/question/1695370

#learnwithBrainly

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Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of
sweet [91]

Answer:

1.44\times 10^{-3} N

Explanation:

We are given that three charged particle are placed at each corner  of equilateral triangle.

q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC

q_1=-8.2\times 10^{-9} C

q_2=-16.4\times 10^{-9} C

q_3=8.0\times 10^{-9} C

Side of equilateral triangle =3.3 cm=\frac{3.3}{100}=0.033m

We know that each angle of equilateral angle=60^{\circ}

Net force=F =\sum\frac{kQq }{d^2}

Where k=9\times 10^9 Nm^2/C^2

If we bisect the angle at q_3 then we have 30 degrees from there to either charge.

Direction of vertical force  due to charge q_1 and q_2

Therefore, force will be added

Vertical  force=9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})

Vertical net force=9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}

Vertical  force =9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}

Vertical  force=-1.41\times 10^{-3}N (towards q_1}

Horizontal component are opposite in direction then will b subtracted

Horizontal force=9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}

Horizontal force=0.27\times 10^-3} N(towards q_2

Net electric force acting on particle 3 due to particle =\sqrt{F^2_x+F^2_y}

Net force=\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}

Net force=1.44\times 10^{-3} N

3 0
3 years ago
HELP FAST What is voltage, and what is its relationship to amperage and power? <br> (10 points)
Amanda [17]

Answer:

Amperage is the rate of electrical charge per time.

1 amp = 1 coulomb / second

Power is the product of amperage and voltage.

P = IV

Explanation:

6 0
3 years ago
A force of 120 N is applied to the front of a sledge at an angle of 28.00 above the horizontal so as to pull the sledge a distan
PtichkaEL [24]

Answer:

Workdone = 17482.36 Joules

Explanation:

Given the following data;

Force, F = 120 N

Angle, d = 28.0°

Distance, x = 165 m

To find the work done, we would use the following formula;

Workdone = FxCosd

Substituting into the formula, we have;

Workdone = 120*165*Cos(28)

Workdone = 19800 * 0.8830

Workdone = 17482.36 Joules

4 0
3 years ago
Please help! Questions are in the picture
Mars2501 [29]

Answer:

3500j

0.250 kg

14.14 m/s

220500 j

93.91 m/s

Explanation:

K.E =1/2 m v^2

first question

1/2×700×10=3500 j

second question

2×78.2/25^2= 0.250 kg

Third question

8kj=8000j

root (2×8000/80)= 14.14 m/s

Last one

P.E=MGH

so it will be

P.E=50×450×9.8=220500 j

the speed

root (2×220500/50)=93.91 m/s

7 0
3 years ago
What is the magnitude of the momentum of a 11kg object moving at 2.2 m/s?
tangare [24]

Answer:

<h3>The answer is 24.2 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 11 kg

velocity = 2.2 m/s

We have

momentum = 11 × 2.2

We have the final answer as

<h3>24.2 kgm/s</h3>

Hope this helps you

4 0
3 years ago
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