Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Increase in sea water pollution
a. We can calculate the amount of work by calculating the area under the graph.
first area (rectangular): 2.5 x 6 = 15
second area(trapezoid): 1/2 x (6+10) x 2.5 =20
total work done: 35 J
b. the force was first applied = 6 N
F = m.a
a = 6 : 3 = 2 m/s²
vf²=vi²+2as
vf²=6²+2.2.5
vf²=56
vf=7.5 m/s
Answer:
Explanation:
Given:
- time taken by the sun to complete one revolution,
- radial distance of the sunspot,
<u>Therefore, angular speed of rotation of sun:</u>
<u>Now the tangential velocity of the sunspot can be given by:</u>
