The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
The given parameters;
- mass of the ball, m₁ = 0.8 kg
- speed of the ball, u₁ = 2.5 m/s
- mass of the object at rest, m₂ = 2.5 kg
- final velocity of the object at rest, v₂ = 1 m/s
Let the final velocity of the 0.8 kg ball immediately after collision = v₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)
2 = 2.5 + (0.8)v₁
-0.5 = (0.8)v₁

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
Learn more here: brainly.com/question/7694106
Answer:
A) hydrostatic force on top of cube = 882.9N
B) hydrostatic force on sides of cube = 0N
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
<h2>
Answer:</h2>
<em>Hello, </em>
<h3><u>
QUESTION)</u></h3>
<em>✔ We have: KE = PE (potential energy) </em>
<em>PE = m x g x h </em>
The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2
PE1 = PE2 ⇔ PE1/PE2 = 1

The mass m1 is therefore 4 times greater than that of the stone of mass m2.
D. Speed and direction, this is because velocity is a vector quantity so has a magnitude and direction assigned to it because it is the rate of change of displacement.