Answer:
c) 12
Explanation:
A Solar eclipse occurs when The Sun, The Earth and The Moon comes in a straight line with the Moon being in between the Earth and the Sun. At this point the Moon appears to block the Sun and Moon's shadow falls on Earth. This would occur only on the day of the New Moon.
If the Moon's orbit was in the same plane as that of the Earth's orbit. Every new Moon, there would be a Solar Eclipse. The Lunar cycle is of 29.5 Days which means there will be one new Moon every month. So there will be 12 Solar Eclipses every year.
Currently, the orbit of the Moon is tilted at an angle of 5° thus we don't see that many Solar eclipses. Maximum of 5 solar eclipses can occur in an year.
Explanation:
A metal such as copper is a <u>conductor</u> because it provides a pathway for electric charges to move easily. A material such as rubber is an <u>insulator</u> because it <u>resists</u> the flow of electric charges. A material that partially conducts electric current is a <u>semiconductor</u>. These materials include <u>group 3 and group 5</u> elements.
Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length
Therefore,
ΔL = 12*10^-6*(18.5-(-3))*1410 = 0.36378 m
Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)
A. krypton (atomic mass 83.8 amu)
B. argon (atomic mass 39.95 amu)
C. xenon (atomic mass 131.3 amu)
D. radon (atomic mass 222 amu)
hope this helps!