The action-reaction pairs in the given situation are:
- the backpack and the bench
- the and and the wall
<h3>What are action-reaction pairs?</h3>
Action-reaction pairs are two forces which are equal but oppositely directed in their line of action.
Action-reaction pairs are according to Newton's third law of motion.
The action-reaction pairs in the given situation are:
- the backpack and the bench
- the and and the wall
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Answer:
Explanation:
Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.
I = 0.5 mr²
Total kinetic energy, T = 0.5 mv² + 0.5 Iω²
T = 0.5 (mv² + 0.5 mr²ω²)
v = rω
So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²
Rotational kinetic energy is
R = 0.5 Iω² = 0.5 x 0.5 mr²ω²
R = 0.25 mv²
So, R / T = 0.25 / 0.75 = 1/3
Modern observations have shown that the geometry of the universe is flat and the universe mist be infinite.
<h3>What is geometry?</h3>
Geometry refer to a branch of mathematics that encapsulates measurements of lines, shapes, dimensions, points and shapes.
Therefore, Modern observations have shown that the geometry of the universe is flat and the universe mist be infinite.
Learn more about geometry below.
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You would use answer 2O g/cm3
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N