False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Explanation:
4a)the displacement is the distance moved in a direction but since no direction is given, the displacement is equal to the distance
b) the distance moved is 400m because that's the length of the track
The answer would be c.
your equation should look like:
reactants-----> products
Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
= K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
= K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
Directed down