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2 years ago

A sample contains 16 g of a radioactive isotope. how much radioactive

Physics

2 answers:

mafiozo [28]2 years ago

7 0

After one half-life, 8 g of radioactive isotope will remain in the sample.

<h3>What is radioactivity?</h3>

The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

The following formula is used to compute the number of half lives elapsed:

$\rm N=\frac{N_0}{2^n} \\\\ N=\frac{16}{2} \\\\ N= 8 \ gram$

Hence,8 gram of radioactive isotope remains in the sample after 1 half-life.

To learn more about the radioactivity, refer to the link;

brainly.com/question/1770619

#SPJ1

Basile [38]2 years ago

5 0

Answer:

Answer is below

Explanation:

8 g

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3 years ago

What two things are a included when describing force? Question 3 options: Direction and mass Strength and size Size and directio

Vinvika [58]

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3 years ago

.A 0.2-kg aluminum plate, initially at 20°C, slides down a 15-m-long surface, inclined at a 30 angle to the horizontal. The forc

Ainat [17]

To solve this problem, it is necessary to apply the concepts related to Work according to the Force and distance, as well as the concepts related to energy lost-or gained-by heat. Mathematically the energy corresponding to heat is given as:

$Q = mC_p(\Delta T)$

Where,

m = mass

$C_p$ = Specific heat

$\Delta T$ = Change in Temperature

At the same time the Work made by the Force and the distance is given as:

$W = F*d \rightarrow W=mg*d$

As the force is applied at an angle of 30 degrees, the efficient component would be given by the vertical then the work / energy would be determined as:

$W = mg*dsin(30)$

$W = (15m)(0.2kg)(9.81)(sin30)$

$W = 13.24J$

Now this energy is used to heat the aluminum. We can find the change at the temperature as follow:

$Q = mC_p(\Delta T)$

$13.24 = (0.2)(900)(\Delta T)$

$\Delta T = 0.0736 \°C$

Therefore the correct answer is B.

7 0

3 years ago

A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles

slamgirl [31]

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035 × v² + 1/2 × 40 × 0.012²

0.032 - 1/2 × 40 × 0.012² = 1/2 × 0.035 × v²

0.02912 = 1/2 × 0.035 × v²

1/2 × 0.035 × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm, v ≈ 1.28996 m/s

E = PE + KE

0.032 = 1/2 × 40 × 0.025² + KE

KE = 0.032 - 1/2 × 40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.

4 0

3 years ago

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grandymaker [24]

Answer:

tssths

Explanation:

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8 0

3 years ago