Answer:
k = 49 N/m
Explanation:
Given that,
Mass, m = 250 g = 0.25 kg
When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.
The force due to mass is balanced by its weight as follows :
mg=kx
So, the spring constant of the spring is 49 N/m.
Answer:
hello your question is incomplete attached below is the complete question
answer :
a) I1 = I2
b) J1 > J2
c) E 1 > E2
d) ( vd1 ) > ( vd2 )
Explanation:
a) The currents in the two segments are the same i.e. I1 = I2 and this is because the segments are connected in series
b) Comparing the current densities J1 and J2 in the two segments
note : current density ∝ 1 / area
The area of the second segment is > the area of first segment therefore
J1 > J2
J1 ( current density of first segment )
J2 ( current density of second segment )
c) Comparing the electric field strengths E1 and E2
note : electric field strength ∝ current density
since current density of first segment is > current density of second segment and conductivity of the materials are the same hence
E 1 > E2
d) Comparing the drift speeds Vd1 and Vd2
( vd1 ) > ( vd2 )
this because ; vd ∝ current density
