Acrow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. what is the kinetic energy of the clam when it is 5.0m above the ground? what is its speed at that point?

The kinetic energy of the clam at a height of 5.0 m is 5.19 J and the speed of the clam at that height is 9.71 m/s.

Explanation:

Mechanical energy is constant throughout the travel, we know that mechanical energy is calculated by adding potential energy and kinetic energy. Potential energy = , Kinetic energy = and Mechanical energy = Kinetic energy is zero at initial point. Now mechanical energy of clam with m=0.11kg,g=9.81,h=9.8 m is = 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of clam at a height of 5.0 m = =. We know that mechanical energy is constant hence, mechanical energy of clam at height 9.8 m is equal to mechanical energy at height 5.0 m. This is represented as following

10.58 = 10.58 – 5.39 =  5.19 = kinetic energy of the clam is 5.19 J.

Now speed of the clam at height 5.0 m is 5.19 = 94.36 = = 9.71 m/s. The speed of the clam is 9.71 m/s.

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