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artcher [175]
3 years ago
5

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

Physics
1 answer:
Elenna [48]3 years ago
8 0
Let the ball is initially at point P1. Finally it reaches at point P2.( bottom most). Since only conservative force is acting on the ball throughout the motion, mechanical energy will be conserved. Therefore conserving energy from P1 to P2, Initially ball is at rest(P1). So initial kinetic energy is 0. Let ball is at height h1 initially, so initial potential energy = mgh ( m= mass of the ball). At bottom most point (P2), potential energy of ball changes into kinetic energy. So final kinetic energy =1/2 mv^2. Final potential energy = 0. Equating energies at P1 and P2, Mgh= 1/2 mv^2. Therefore, v= (2gh)^1/2. Since all the initial and final points are at same height, velocity will be same for all the paths(irrespective of the path taken by ball). So velocity will be same.
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A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.

The most correct answer is:

* Propagation of pressure fluctuations in a medium

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3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

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After reading this whole question, I feel like I've already
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