(a) 392 N/m
Hook's law states that:
(1)
where
F is the force exerted on the spring
k is the spring constant
is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:

- The stretching of the spring is

Solving eq.(1) for k, we find the spring constant:

(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is

And so, the stretch of the spring is

And since the initial lenght of the spring is

The final length will be

<h3><u>Question: </u></h3>
The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?
a. The mass of the sun
b. The mass of the satellite
c. The mass of the Earth
<h3><u>Answer:</u></h3>
The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.
Option c
<h3><u>
Explanation:
</u></h3>
Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence
.
Gravitational force between Earth and Satellite: 
Centripetal force of Satellite :
Where G = Gravitational Constant
= Mass of Earth
= Mass of satellite
R= Radius of satellite’s circular orbit
V = Speed of satellite
Equating
, we get
Speed of Satellite 
Thus the speed of satellite depends only on the mass of Earth.
B - A theory seems to be the closest
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