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Ad libitum [116K]
3 years ago
14

If you know how much radioactive material the organism had to begin with, explain how you could use half-life to determine its a

ge now.
Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer: We can calculate it with the radioactive half life equation

Explanation:

If we already know the initial amount of radioactive material and its half life, we can leave that material for a specific known time and then measure how much of the material is left (since it follows the radioactive deacay) and use the results in the following formula:

A=A_{o}.2^{\frac{-t}{h}}  

Where:

A is the final amount of the material

A_{o} is the initial amount of the material

t is the time elapsed

h is the half life of the radioactive compound

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What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?
madam [21]

Answer:

there are 25 kg objective travelling at 2m/s to the right.

4 0
2 years ago
You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p
user100 [1]

Answer:

The potential energy at point A is 17.1675 J

Explanation:

The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries

In meteorology it is used to describe the level of saturated soil above the water table

Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are

Point A, elevation = 75 cm capillary potential = -100 cm

Point B, elevation = 25 cm capillary potential = -200 cm

The total potential energy at point A is

Elevation above reference - capillary potential =75-(-100) = 175 cm

which gives per unit mass

PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J

8 0
4 years ago
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b
Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

Q=t^3-2t^2+4t+4

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4

At t = 1 s,

Current,

I=3(1)^2-4(1)+4\\\\I=3\ A

So, the current at t = 1 s is 3 A.

For lowest current,

\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s

Hence, this is the required solution.

7 0
4 years ago
We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi
antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

Kc=510=1/2*Ic*\omega c^2

\omega c=\sqrt{510*2/Ic}

Where the inertia is given by:

Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2

Replacing this value:

\omega c=106.46rad/s

Speed of the block will therefore be:

V_b=\omega_c*R_c=106.46*0.15=15.969m/s

By conservation of energy:

Eo = Ef

Eo = 0

Ef = 510+1/2*m_b*V_b^2-m_b*g*h

So,

0 = 510+1/2*m_b*V_b^2-m_b*g*h

Solving for h we get:

h=16.67m

3 0
4 years ago
Of all the radio waves, these have the shortest wavelengths and the the highest frequency?
Julli [10]
If by chance you mean any wave, the wave with the shortest wavelength/highest frequency is gamma rays.
4 0
3 years ago
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