Answer:
It is used in MRI because it does not damage cells
Radio waves are used for space research because they have very long wavelengths
Explanation:
Many parts of the electromagnetic spectrum are applied in clinical diagnosis and treatment of illnesses. However, these highly ionizing radiation damage cells and its dosage must be carefully managed to avoid creating radiation related health problems for the patients.
Radio waves can be used in MRI without issues because the energy of the radiation is not sufficient to cause damage to cells but is sufficient to provide images for the sake of medical diagnosis.
Secondly, radio waves have long wavelength. This property is suitable for long range
communication. Hence it can be used in space research
Answer:
by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products. - This is the answer.
Explanation:
Answer:
9 terms. In carbon dioxide (CO2), there are two oxygen atoms for each carbon atom. Each oxygen atom forms a double bond with carbon, so the molecule is formed by two double bonds. Two double bonds means that the total number of electrons being shared in the molecule is.
Explanation:
Answer:
gas is dioatomic
T_f = 330.0 K

Explanation:
Part 1
below equation is used to determine the type Gas by determining
value

where V_i and V_f is initial and final volume respectively
and P_i and P_f are initial and final pressure


\gamma = 1.38
therefore gas is dioatomic
Part 2
final temperature in adiabatic process is given as
](https://tex.z-dn.net/?f=T_f%20%3D%20T_i%2A%5B%5Cfrac%7Bv_i%7D%7BV_f%7D%5D%28%5E%5Cgamma-1%29)
substituing value to get final temperature
![T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}](https://tex.z-dn.net/?f=T_f%20%3D%20260%2A%5B%5Cfrac%7B151%7D%7B80.6%7D%5D%5E%20%7B%281.38-1%29%7D)
T_f = 330.0 K
Part 3
determine number of moles by using following formula



Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that the speed of light is given as


now the frequency of the light is given as

so we have


Part b)
Position of Nth maximum intensity on the screen is given as

so here we know for 3rd order maximum intensity

n = 3
L = 1.4 m


Part c)
angle of third order maximum is given as


