Answer:
a) the magnitude of the force is
F= Q(
) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = 
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = 
, where p = q × s
E(r) = 
where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
Scientific Notation: 3.45 x 10^5
E Notation: 3.45e5
<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.
<h3>What is a momentum?</h3>
- In Newtonian physics, an object's linear momentum, translational momentum, or simply momentum is defined as the product of its mass and velocity.
- It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
- The kilogram metre per second (kg m/s), or newton-second in the International System of Units (SI), is the unit used to measure momentum.
- The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion.
To know more about momentum, refer:
brainly.com/question/1042017
#SPJ4
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force 
Answer:
all of the above
Explanation:
- a build up of electric charge
- the force and motion of electrically charged particles
- an electric current
are three different ways to describe electricity.
So the answer is all of the above.
