Lets se
And


So

If spring constant is doubled mass must be doubled
Answer:
Electric field acting on the electron is 127500 N/C.
Explanation:
It is given that,
Mass of an electron, 
Charge on electron, 
Initial speed of electron, u = 0
Final speed of electron, 
Distance covered, s = 2 cm = 0.02 m
We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :



According to Newton's law, force acting on the electron is given by :
F = ma


Electric force is given by :
F = q E, E = electric field


E = 127500 N/C
So, the electric field is 127500 N/C. Hence, this is the required solution.
Answer:
C) W = - 190 J
Explanation:
Notation
Wf = work done by the friction force (unknown)
Ff = force of the friction
d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S
= 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S
> σ
Answer: The changing magnetic field caused by the material's motion induces a current in the coil of wire proportional to the change in field. If a 0 is represented, the magnetic field does not change between the two domains of a bit, so no current is induced as the magnetic material passes the coil.