The force between the two point charge when they are separated by 18 cm is 3 N
<h3>How do I determine the force when they are 18 cm apart?</h3>
Coulomb's law states as follow:
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ => constant
F₁r₁² = F₂r₂²
Where
- F₁ and F₂ are the initial and new force
- r₁ and r₂ are the initial and new distance apart
With the above formula, we can obtain the force between the two point charge when they are 18 cm apart. Details below:
- Initial distance apart (r₁) = 6 cm
- Initial force of attraction (F₁) = 27 N
- New distance apart (r₂) = 18 cm
- New force of attraction (F₂) =?
F₁r₁² = F₂r₂²
27 × 6² = F₂ × 18²
972 = F₂ × 324
Divide both side by 324
F₂ = 927 / 324
F₂ = 3 N
Thus, the force when they are 18 cm apart is 3 N
Learn more about force:
brainly.com/question/28569085
#SPJ1
I don't know what the tables you have look like but her <span>displacement</span> would be -1m
Answer: C. 1.4 10-11 N up
Explanation:
The magnetic force, F on a charge q moving with velocity v in a magnetic field B at an angle θ is given by:
F = q v B sin θ
Charge of proton, q = 1.6 × 10⁻¹⁹ C
Strength of magnetic field, B = 3.4 T pointing outwards
velocity of the proton, v = 2.5 × 10⁷ m/s towards left
Magnetic force is given by:
F = 1.6 × 10⁻¹⁹ C× 2.5 × 10⁷ m/s ×3.4 T× sin 90 = 13.6 × 10⁻¹² N = 1.4 × 10⁻¹¹ N up
The direction of the force is given by Lorentz Right hand rule. The fingers point magnetic field, the thumb points towards velocity, then the force on the proton is given by the direction perpendicular to the palm.
The magnetic field acts outwards with velocity of the proton towards left. The force would act perpendicular to the two -upwards.
Answer:
4 hydrogen atoms can form chemical bond with 1 carbon atom.
Explanation:
CH4. methane
Answer:
1.1259*10^9 Newton per Columb
Explanation:
the magnitude of the electric field intensity can be calculated using the expresion below;
E=Kq/r^2
Where k= constant
q= electric charge
r=distance= 2cm= 20*10^-2m( we convert to m for unit consistency
:,K=59*10^9 Columb
If we substitute the value into above formula we have
E=(9*10^9)*(5*10^-3)/(20*10^-2)^2
=1.1259*10^9 Newton per Columb
Therefore,the magnitude of the electric field intensity in vacuum at a distance of 20 cm is 1.1259*10^9 Newton per Columb
