Answer:
The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.
ANSWER:
The answer will be OT
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
This electric force calculator will enable you to determine the repulsive or attractive force between two static charged particles. Continue reading to get a better understanding of Coulomb's law, the conditions of its validity, and the physical interpretation of the obtained result.
How to use Coulomb's law
Coulomb's law, otherwise known as Coulomb's inverse-square law, describes the electrostatic force acting between two charges. The force acts along the shortest line that joins the charges. It is repulsive if both charges have the same sign and attractive if they have opposite signs.
Coulomb's law is formulated as follows:
F = keq₁q₂/r²
where:
F is the electrostatic force between charges (in Newtons),
q₁ is the magnitude of the first charge (in Coulombs),
q₂ is the magnitude of the second charge (in Coulombs),
r is the shortest distance between the charges (in m),
ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C². This value is already embedded in the calculator - you don't have to remember it :)
Simply input any three values
Answer:
1. Why is Jupiter's rotation dangerous for human survivability?
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<em><u>Jupiter is the fastest rotating planet in our solar system. One day lasts about 9.5 Earth hours. This creates powerful winds that can whip around the planet at more than 300 mph. About 75 miles below the clouds, you reach the limit of human exploration.</u></em>
2 .Why is Jupiter's planet axis tilt an issue for human survivability?
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<em><u>Jupiter, like Venus, has an axial tilt of only 3 degrees, so there is literally no difference between the seasons. ... The length of each season is roughly three years. Jupiter is the fastest spinning planet in our Solar System, which causes the planet to flatten at the poles and bulge at the </u></em><em><u>equator.</u></em>
3.Why is the diameter of Jupiter an issue for human survivability?
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<em><u>Since </u></em><em><u>,</u></em><em><u>The </u></em><em><u>Jupiter </u></em><em><u>is </u></em><em><u>so </u></em><em><u>huge </u></em><em><u>in </u></em><em><u>mass</u></em><em><u> </u></em><em><u>,</u></em><em><u>The </u></em><em><u>central</u></em><em><u> </u></em><em><u>force</u></em><em><u> </u></em><em><u>toward</u></em><em><u> </u></em><em><u>the </u></em><em><u>centre </u></em><em><u>will </u></em><em><u>be </u></em><em><u>high</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>we'll</u></em><em><u> </u></em><em><u>be </u></em><em><u>forced</u></em><em><u> </u></em><em><u>toward</u></em><em><u> </u></em><em><u>it </u></em><em><u>causing</u></em><em><u> </u></em><em><u>Several</u></em><em><u> </u></em><em><u>problems</u></em><em><u>.</u></em>