Answer: 0.333 h
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula</u>:
(1)
Where:
is the final amount of the material
is the initial amount of the material
is the time elapsed
is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find
from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Clearing
:
(6)
Finally:
This is the half-life of the Bismuth-218 isotope
Answer:
The strength of the magnetic field that the line produces is
.
Explanation:
From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:
(1)
Where
is the permiability constant, I is the current and r is the distance from the wire.
Notice that it is necessary to express the current, I, from kiloampere to ampere.
⇒ 
Finally, equation 1 can be used:
Hence, the strength of the magnetic field that the line produces is
.
According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so



PART B) Replacing the values given as,




Therefore the speed of the masses would be 1.8486m/s
Answer:

Explanation:
Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.
This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

Where:

Solving for λ:

Replacing the data provided by the problem:

Answer:
When a ray of light passes through a glass slab of a certain thickness, the ray gets displaced or shifted from the original path. This is called lateral shift/displacement.
Explanation:
.