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Answer:
<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>
Explanation:
<u>Net Force
</u>
The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.
The acceleration is proportional to the net force and inversely proportional to the mass of the object.
If the object has zero net force, it won't get accelerated and its velocity will remain constant.
The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.
Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.
The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.
Answer:
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Answer:
10mm
Explanation:
According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.
Mathematically, F = ke where;
F is the applied force
k is the elastic constant
e is the extension
from the formula k = F/e
k = F1/e1 = F2/e2
Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm
Let F1 = 1N e1 = 2mm
The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?
Substituting this values in the formula above we have
1/2=5/e2
Cross multiplying;
e2 = 10mm
This shows that we must have dent it by 10mm before it pushes outwards by a 5N force
Answer:
31.8 × 10⁻⁴ J = 3.18 mJ
Explanation:
We know the intensity I of a wave is I = P/A where P = power and A = area = 0.500 m²
The intensity of an electromagnetic wave is also equal to I = E₀²/μ₀c
where E₀ = maximum electric field strength = √2E where E = rms value of electric field = 0.0200 N/C, μ₀ = 4π × 10⁻⁷ H/m ,c = 3 × 10⁸ m/s
P/A = E₀²/μ₀c = 2E²/μ₀c
P = 2E²A/μ₀c = 2 × (0.02 N/C)² × 0.5 m²/(4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)
= 1.06 × 10⁻⁴ W = 0.106 mW
Since P = E/t where E = Energy and t = time
E = Pt with t = 30 s
E = 1.06 × 10⁻⁴ W × 30 s = 31.8 × 10⁻⁴ J = 3.18 mJ
So the wave carries 3.18 mJ of energy through the window in 30 s
