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kkurt [141]
4 years ago
11

1. What happens in the process of radioactive decay? What is the half-life of a radioactive substance, and how is it used to dat

e an object?
Physics
1 answer:
Ierofanga [76]4 years ago
6 0
<span> In radioactive decay, an unstable atomic nucleus emits particles or radiation and converts to a different atomic nucleus. If the new nucleus is unstable, it will decay again, until eventually, a stable nucleus is formed. Such a sequence of nuclear decays forms a decay series.

The half-life of a radioactive substance is the time required for half of the atoms of a radioactive isotope to decay. If you have, say, 1 million atoms of a specific isotope in a sample, the time required for 500,000 of those atoms to decay is the half-life of that specific isotope. If you have 50 atoms of that isotope, 25 atoms will decay in the same amount of time.

Because the half-life is fixed for a specific isotope, it can be used to date objects. You compare the decay rate of an old object with the decay rate of a fresh sample. Nuclear decay is a first-order process and can be described by a specific mathematical equation, which depends on the decay rate and the half-life. Knowing those values, you can work back and determine the age of an object, as compared with a standard sample. Old objects will not have as much of a radioactive isotope in them as new objects, since the isotopes will have decayed over time in the old object.</span>
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Will mark as BRAINLIEST....... The Displacement x of particle moving in one dimension under the action of constant force is rela
pentagon [3]

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

x=4t^3+3t^2-5t+2

Where,

x is in meters and t is in sec

We know that,

Velocity,

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5

(a) i. t = 2 s

v=12(2)^2+6(2)-5=55\ m/s

At t = 4 s

v=12(4)^2+6(4)-5=211\ m/s

(b) Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6

Pu t = 3 s in the above equation

So,

a=24(3)+6\\\\a=78\ m/s^2

Hence, this is the required solution.

3 0
3 years ago
Begin any simulation, and turn on Gravity Force in the central menu on the right. The gravity force arrow shows the direction an
Katyanochek1 [597]

Answer:

The gravity arrow for each body rotates, always pointing toward the other body. Both arrows grow longer when the bodies come closer to one another and shorter when they move farther apart. This change shows that the gravitational force is stronger the closer together the bodies are.

Explanation:

3 0
3 years ago
A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment
snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = I\alpha

\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2

Now

\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians

Here t = 1 minutes = 60 seconds

3 0
3 years ago
What would you use to to measure the weight of a truck?? I’m needing a word that is 8 letters long
myrzilka [38]

Answer:

Railroad Scale?

Explanation:

5 0
3 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
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