Answer:

0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that

Where 

u(0)=0
Substitute the value

Substitute u'(0)=10


Substitute the values

Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec
Answer
given,
mass of the ski = 75 Kg
speed of the skier, v = 3 m/s
time = 1.50 min = 90 s
angle of inclination, θ = 40°
distance = s x t
= 3 x 90 = 270 m
a) W = F. d cos θ
W = mg. d cos θ
W = 75 x 9.8 x 270 x cos 40°
W = 152021.52 J
work is done by the ski lift is equal to 152021.52 J
b) Power extended by the ski


P = 1689.13 Watt.
power is expended by the ski lift is equal to 1689.13 W.
Answer:
Explanation:
Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km
Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km
Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite
Orbital potential energy of a satellite A = - GMm / Ra
Orbital potential energy of a satellite B = - GMm / Rb
PE of satellite B /PE of satellite A
= Ra / Rb
= 12740 / 25480
= 1 / 2
b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same
KE of satellite B /KE of satellite A
= 1 / 2
c ) Total energy will be as follows
Total energy = - PE + KE
- P E + PE/2
= - PE /2
Total energy of satellite B / Total energy of A
= 1 / 2
Satellite B will have greater total energy because its negative value is less.
The first one might be faunal succsession and the 2nd one might be metamorphic rock