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
Answer:
the speed of the waves is 150 cm/s
Explanation:
Given;
frequency of the wave, f = 10 Hz = 10
distance between 4 nodes, L = 15.0 cm
The wavelength (λ) of the wave is calculated as follows;
Node to Node = λ/2
L = 2(Node to Node) = (4 Nodes) = 2 (λ/2) = λ
Thus, λ = L = 15.0 cm
The speed (v) of the wave is calculated as follows;
v = fλ
v = 10 Hz x 15.0 cm
v = 150 cm/s
Therefore, the speed of the waves is 150 cm/s
Answer:

Explanation:
The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:


Now, the exerted force is:

The volume of a sphere is:



Lastly, the force is calculated:


An object with volume greater than it mass. to account less density and thus it will float