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 mass of the man is 71 kg
Explanation:
Given;
kinetic energy of the man, K.E = 887.5 J
velocity of the man, v = 5 m/s
The mass of the man is calculated as follows;
K.E = ¹/₂mv²
where;
m is the mass of the man
2K.E = mv²
m = 2K.E / v²
m = (2 x 887.5) / (5)²
m = 71 kg
Therefore, the mass of the man is 71 kg
Explanation:
see the above attachment to solve the question and get the answer.
hope this helps you.
As we know that two charges exert force on each other when they are placed near to each other
The force between two charges is given as
here we know that
= two different point charges
r = distance between two point charges
also we know that two similar charges always repel each other while two opposite charges always attract each other
so here correct answer would be
<em>A. A positive and negative charge attract each other.</em>
