Answer: 6.9x 107 in standard form is 69,000,000
Answer:
(a) Length =136.58 m
(b) T=5995 N
Explanation:
for the glider in the back
T - 1900 = 700 a
for the glider in front
12000-T -1900 = 700a
add equations
12000-3800 = 1400 a
a=5.85 m/s^2
v^2 = v0^2 + 2 a x
40^2 = 2*5.85*x
Length =136.58 m
b) plug the a back into one of the previous formula
T - 1900 = 700*5.85
T=5995 N
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
Explanation:
Fgravity = G*(mass1*mass2)/D²
so, if you double one of the masses, what does that do to our equation ?
Fgravitynew = G*(2*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity
so, the correct answer will be 2×45 = 90 units.
Answer:
The distance between the two spheres is 914.41 X 10³ m
Explanation:
Given;
4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;
1 e = 1.602 X 10⁻¹⁹ C
4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C
V = Ed
where;
V is the electrical potential energy between two spheres, J
E is the electric field potential between the two spheres N/C
d is the distance between two charged bodies, m

where;
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063
d = 914.41 X 10³ m
Therefore, the distance between the two spheres is 914.41 X 10³ m