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Elodia [21]
3 years ago
5

What is the half-life of a radioisotope

Physics
2 answers:
IceJOKER [234]3 years ago
4 0
The rate at which a radioactive isotope<span> decays is measured in </span>half-life. The termhalf-life<span> is defined as the time it takes for one-</span>half<span> of the atoms of a radioactive material to disintegrate. </span>Half-lives<span> for various </span>radioisotopes<span> can range from a few microseconds to billions of years.</span>
HACTEHA [7]3 years ago
4 0

Answer:

It is the time after which a given sample of radioactive substance will decay to half of the initial number present in the sample.

Explanation:

As we know that as per law of radioactivity the number of radioactive nuclei present in the sample is proportional to the rate of decay

so it is given by

\frac{dN}{dt} = -\lambda N

after solving above relation we will have

N = N_0e^{-\lambda t}

now here we know that as the number of nuclei is reduced to half of the initial number then we have

\frac{N_0}{2} = N_0e^{-\lambda t}

t = \frac{ln2}{\lambda}

so this time is known as half life where the denominator of the equation is known as decay constant.

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A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall
Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object m=2\ kg

Speed of object u=5\ m/s

Mass of object at rest M=3\ kg

Suppose after collision, speed is v

conserving momentum

\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s

Initial kinetic energy

k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J

Final kinetic energy

k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

4 0
3 years ago
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s
dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0
3 years ago
The work done on a box is 532 joules. The force applied to the box was 48 N. What was the displacement of the box? *
Semenov [28]

Explanation:

Work = force × displacement

532 J = 48 N × d

d ≈ 11 m

8 0
3 years ago
When a burning stick of increase is moved fast in a circle a circle of red light is seen.​
anzhelika [568]

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

7 0
3 years ago
If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d
azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt     from this equation we clear "is" = fs - gt

Substitution                         is = 77 - (9,81)(6.5)

Process                               is = 77 - 63.8

                                            is = 13.2 m/s

8 0
3 years ago
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