Answers
Answer:
The new force is 2/3 of the original force
Explanation:
Coulomb's Law
The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Written as a formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:
[tex]\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}[/tex]
Factoring out 2/3:
[tex]\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}[/tex]
Substituting the original force:
[tex]F'=\frac{2}{3}F[/tex]
The new force is 2/3 of the original force
Related Questions
A wheel of mass 50 kg has a radius of 0.4 m. It is making 480 rpm. What is the
torque necessary to bring it to rest in 40 seconds?
Solution:
Answers
Answer:
The torque necessary to bring the wheel to rest in 40 seconds is 10.4 N·m
Explanation:
The question is with regards to rotational motion
The rotary motion parameters are;
The mass of the wheel = 50 kg
The radius of the wheel = 0.4 m
The rate of rotation of the wheel = 480 rpm
The time in which the wheel is to be brought to rest = 40 s
The rotational rate of the wheel in rotation per second is given as follows;
480 r.p.m = 480 r.p.m × 1 minute/(60 seconds) = 8 revolution/second
1 revolution = 2·π radians
Therefore, we have the angular velocity, ω, given as follows;
ω = 2·π × 8 revolutions/second ≈ 50.3 rad/s
The angular acceleration, α, is given as follows;
[tex]\alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{\omega _2 - \omega_1}{t_2 - t_1}[/tex]
Whereby the wheel is brought to rest from its initially constant rotational motion in 40 seconds, we have;
ω₁ ≈ 50.3 rad/s, ω₂ = 0 rad/s, and t₂ - t₁ = 40 seconds
Plugging in the values for the variables of the equation for the angular acceleration, "α", we get;
[tex]\alpha = \dfrac{0 - 50.3 \ rad/s}{40 \ s} \approx 1.3 \ rad/s^2[/tex]
The torque on the wheel, τ, is given as follows;
τ = m·r²·α
Where;
m = The mass of the object = 50 kg
r = The radius of the wheel = 0.4 m
α = The acceleration of the wheel ≈ 1.3 rad/s²
Therefore;
τ = 50 kg × (0.4 m)² × 1.3 rad/s² ≈ 10.4 N·m
The torque necessary to bring the wheel to rest in 40 seconds = τ ≈ 10.4 N·m.
Answer:
-10.048 N m
Explanation:
Compounds are made from the atoms of two or more______?
Answers
Answer:
elements
not really an explanation