A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = m v²

=   x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = = 235 N

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

laticia draws the diagram below to show earth’s magnetic field.

answer

the arrows to the right of earth should point up.