In Newton's third law, the action and reaction forces D.)act on different objects
Explanation:
Newton's third law of motion states that:
<em>"When an object A exerts a force on object B (action force), then action B exerts an equal and opposite force (reaction force) on object A"</em>
It is important to note from the statement above that the action force and the reaction force always act on different objects. Let's take an example: a man pushing a box. We have:
- Action force: the force applied by the man on the box, forward
- Reaction force: the force applied by the box on the man, backward
As we can see from this example, the action force is applied on the box, while the reaction force is applied on the man: this means that the two forces do not act on the same object. This implies that whenever we draw the free-body diagram of the forces acting on an object, the action and reaction forces never appear in the same diagram, since they act on different objects.
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Answer:
Carbon dioxide and water
Explanation:
The products of complete combustion are always carbon dioxide and water.
The balanced reaction is:
4 CH₃OH + 3 O₂ → 4 CO₂ + 2 H₂O
Answer:
N= 3
Explanation:
For this exercise we must use Faraday's law
E = - dФ / dt
Ф = B . A = B Acos θ
tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives
E = - A cos θ (B - B₀) / t
The angle enters the magnetic field and the normal to the area is zero
cos 0 = 1
A = π r²
In the length of the wire there are N turns each with a length L₀ = 2π r
L = N (2π r)
r = L / 2π N
we substitute
A = L² / (4π N²)
The magnetic field produced by a solenoid is
B = μ₀ N/L I
for which
B₀ = μ₀ N/L I
The final field is zero, because the current is zero
B = 0
We substitute
E = - (L² / 4π N²) (0 - μ₀ N/L I) / t
E = μ₀ L I / (4π N t)
N = μ₀ L I / (4π t E)
The electromotive force is E = 0.80 mV = 0.8 10⁻³ V
let's calculate
N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]
N = 320 10⁻⁷ / 9.6 10⁻⁶
N = 33.3 10⁻¹
N= 3
Answer: 1.176×10^-3 s
Explanation: The time constant formulae for an RC circuit is given below as
t =RC
Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F
t = 56×10^-6 × 21
t = 1176×10^-6
t = 1.176×10^-3 s