Three units deformations observed in these bands.
<h3>What forces do a rubber band encounter?</h3>
Elastic force is the force that permits some materials to regain their former shape after being stretched or crushed. Thus, a stretched rubber band is subject to an elastic force.
The rubber band experiment uses a straightforward rubber band to show entropic force and a refrigeration cycle. The rubber band experiment involves stretching and then releasing a rubber band while measuring its temperature.
Always acting in the opposite direction of motion is friction. This indicates that if friction is there, it cancels out some of the force driving the motion (if the object is being accelerated). This results in a decreased acceleration and a smaller net force.
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Answer:
Each part so obtained will represent the fraction 1/8 and the number line obtained will be of the form: To mark 3/8; move three parts on the right-side of zero. To mark 5/8; move five parts on the right-side of zero. To mark -1 3/8 i.e. -11/8; move eleven parts on the left-side of zero.
Explanation:
Distance = 1/2*gravity*velocity^2
<span>So plugging: 1/2 * 9.8 * 25 = 122.5units
and is six second
</span><span>Distance = 1/2*gravity*velocity^2
</span><span>So plugging: 1/2 * 9.8 * 36= 176.4 units</span>
Answer:
a. before
Explanation:
Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?
from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .
displacement is distance in a specific direction.
before the displacement was maximum at 2mm was instant at time=0.04s.
But later was constant at 0.06s at a displacement point of 1mm
Answer:
The new intensity = 3.38 × 10⁻⁵ W/m²
Explanation:
Intensity of sound wave:
The intensity of sound is the rate of flow of flow of energy, per unit area, perpendicular to the direction of the sound wave.
Intensity (I) ∝ A²
where I = intensity, A = Amplitude.
∴ I₁/I₂ = A₁²/A²₂............................... equation 1
From the question, the amplitude increase by 30% of the initial
∴ A₂ = A₁ + 0.3A₁ = 1.3A₁, I₁ = 2.00×10⁻⁵ W/m²
∴ (2.00×10⁻⁵)/I₂ = A₁²/(1.3A₁)²
(2.00×10⁻⁵)/I₂ = 1/1.3²
making I₂ the subject of the equation
I₂ = (2.00 × 10⁻⁵)×1.3² = 3.38 × 10⁻⁵ W/m₂
The new intensity = 3.38 × 10⁻⁵ W/m²
