Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Lighter molecules move fast and escape from the upper atmosphere relatively quickly.
To find the answer, we have to know more about the lighter isotopes.
<h3>
What are lighter isotopes?</h3>
- Lighter molecules are mobile and soon leave the higher atmosphere.
- A particular element's stable isotopes have slightly different atomic masses and quantum mechanical energies.
- The lighter isotope of an element's chemical bonds are more easily broken than the heavier isotope's.
- As a result, the light isotope typically benefits from chemical reactions.
Thus, we can conclude that, lighter molecules move fast and escape from the upper atmosphere relatively quickly.
Learn more about the isotopes here:
brainly.com/question/364529
#SPJ4
1. All of the above
2. Lack of consistent sunlight
3. Nonpolluting & Can be developed anywhere
4. Protons and Neutrons
sorry the answer is so late...Hope it still helps u
:)
Also I'm pretty sure all the answers I provided are correct but I'm not sure 4 a fact so plz let me know...
Answer:
16 degrees
Explanation:
The tipping point of the cabinet is sketched below.
