Answer:
E=7453.99 V/m
Explanation:
The electric field on the charged is given by
E= Kqx/(r^2 +x^2)^3/2
Where;
K= constant of Coulomb's law
q= magnitude of charge= 30.0×10^-9 C
r= radius of the rings= 5 cm or 0.05m
x= distance between the rings = 18cm = 0.18 m
Substituting values;
E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2
E= 48.6/(2.5×10^-3 + 0.0324)^3/2
E= 48.6/(0.0025 + 0.0324)^3/2
E= 48.6/6.52×10^-3
E=7453.99 V/m
Planets orbit the sun in the paths which are known as elliptical orbit. Each planet has its own orbit around the sun and direction in which all the planets orbit around the sun are the same. These orbits were well explained by the astronomer Kepler. The gravity of the Sun keeps the planets in their orbits. They stay in their orbits because there is no other force in the Solar System which can stop them.
Yes it is possible only if the x axis is distance and the y axis is time, providing the object being monitored or graphed never moves and you are utilizing a line graph for the assignment. time will increase (traveling vertically on the y axis or up/down on the graph) and distance will stay at a constant 0 (not traveling horizontally the x axis or left/right on the graph)
1. To solve this problem, you must apply formula of Universal Gravitation Law, which is shown below:
F=Gm1m2/r²
F=2.75x10^-12
G=6.7x10^-11
r=2.6 m
m2=2m1
2. You must clear m1, as below:
F=G(m1)(2m1)/r²
F=G(2m1)²/r²
m1=√(Fr²/2G)
3. When you susbtitute the values into the formula m1=√(Fr²/2G), you obtain:
m1=√(Fr²/2G)
m1√0.13
m1=0.37
m2=2m1
m2=2(0.37)
m2=0.74
4. Therefore, the answer is:
m1=0.37
m2=0.74