Answer:
The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."
Explanation:
The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame
take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2
relative velocity of the object to the S' frame would be
Vrel = v2- v
This means the Vrel of the object with respect to the S' frame is less than the Vrel of the object with respect to the S frame
However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.
This would mean the second option is the answer, the relative speed of the object depends on the actual values.
<span>The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .
When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.
The refracted ray makes an angle with the normal known as angle of refraction. The sin of angle of incidence to the sin of angle of refraction is called the refractive index( </span>μ= <span>sin i / sin r) .
hope all of it helps you!</span>
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:
The acceleration of the object decreases I think
Explanation:
