Initially, mg = kx. K = mg/x = 700/0.5x10^-3 = 1400000N/m. From second condition, applying work-energy theorem, potential enery- elastic potential energy = change in kinetic energy. Now change in kinetic energy is 0 since initial and final velocities are 0m/s. Therefore, potential energy = elastic potential energy. mgh = (1/2) * k* x^2. x^2 = 2(mg)h/k = 2 x 700 x 1.3/ 1400000. x = 0.036m. Hope it's clear.
Answer:
magnetic flux linkage includes number of turns, N while magnetic flux has no N
Since number of turns are not significant to units, they become similar.
I found the answer sheet online for you
Answer:
u = 25 m/s
Explanation:
given,
length of skid = 93 m
coefficient of friction = 0.35
final velocity = 0 m/s
initial velocity = ?
force here is friction f = μ mg
F = ma
now com paring
-μ mg = m a
a = - μ g
a = - 0.35 x 9.8
a = -3.43 m/s²
we know,
v² = u² + 2 a s
0 = u² - 2 x 3.43 x 93
u² = 637.98
u = 25.26 m/s
u = 25 m/s (two significant figure)
