The best and most correct answer among the choices provided by the question is <span>B.sound waves</span><span>.
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<span>Particles move together or apart parallel to the direction of the sound wave.
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Answer:
20 ft
Explanation:
This is the length where the direct contact could be used as ground electrode
The cat has two directions of motions:
The horizontal motion = Dx = 2.2 m
The vertical motion = Dy = -1.3 m (negative sign indicates that the cat is falling)
a = 9.8 m/sec^2
Vy = zero (since you are not moving up)
From the laws of motion:
<span>Dy = Vyt + 0.5ayt^2
</span>-1.3 = 0(t) + 0.5(-9.8)t^2
<span>t = 0.52s
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Then, again using the laws of motion (but for the horizontal direction this time)
Dx = Vxt
<span>2.2 = Vx0.52 </span>
<span>Vx = 2.2/0.52 </span>
<span>= 4.23 m/s
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<span>Therefore the cat's speed when it slid off the table is 4.23 m/s horizontally.</span>
Answer:
C. Evaporating water from the container.
Explanation:
The concentration of solution changes when solvent or solute are added/removed from a solution.
Answer:
P=740 KPa
Δ=7.4 mm
Explanation:
Given that
Diameter of plunger,d=30 mm
Diameter of sleeve ,D=32 mm
Length .L=50 mm
E= 5 MPa
n=0.45
As we know that
Lateral strain



We know that




So the axial pressure


P=740 KPa
The movement in the sleeve


Δ=7.4 mm