Answer:
0.2625 N/C
Explanation:
Force of attraction = 
where Q₁ and Q₂ are charges and R is distance between charges.K is a constant equal to 9 x 10⁹.
= 
= 0.2625 N/C
Answer:
F_total = 29.4 N, directed to the right of particle 2
Explanation:
We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.
Let's use Coulomb's law to calculate each force
F = 
particles 1 and 2
q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m
F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²
F₁₂ = 2.59 10¹ N
Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.
particles 2 and 3
q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m
we calculate
F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²
F₂₃ = 3.5 N
as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2
Now we add the forces as vectors
F_total = ∑ F = F₁₂ + F₂₃
F_total = 25.2 +3.5
F_total = 29.4 N
directed to the right of particle 2
The correct answer is A: H3PO4
Answer:
1.13 x 10⁵N
Explanation:
Given parameters:
Pressure of the coin press = 3.2 x 10⁸ Pa
radius of the nickel coin = 0.0106m
Unknown:
Force of the press on coil = ?
Solution:
Our knowledge of pressure will help us solve this problem.
Pressure is defined as the force applied per unit area on a body.
Pressure = 
Force = Pressure x Area
Since the pressure is known;
Area of the coin = Area of a circle = π r²
where r is the radius of the coin;
Area of the coin = π x 0.0106² = 3.53 x 10⁻⁴m²
Force = 3.2 x 10⁸ x 3.53 x 10⁻⁴ = 1.13 x 10⁵N
(a) -2451 N
We can start by calculating the acceleration of the car. We have:
is the initial velocity
v = 0 is the final velocity of the car
d = 125 m is the stopping distance
So we can use the following equation

To find the acceleration of the car, a:

Now we can use Newton's second Law:
F = ma
where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

and the negative sign means the force is in the opposite direction to the motion of the car.
(b) 
We can use again the equation

To find the acceleration of the car. This time we have
is the initial velocity
v = 0 is the final velocity of the car
d = 2.0 m is the stopping distance
Substituting and solving for a,

So now we can find the force exerted on the car by using again Newton's second law:

As we can see, the force is much stronger than the force exerted in part a).