On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.
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Answers

Answer 1
Answer:

Answer:

ω = √(2T / (mL))

Explanation:

(a) Draw a free body diagram of the mass.  There are two tension forces, one pulling down and left, the other pulling down and right.

The x-components of the tension forces cancel each other out, so the net force is in the y direction:

∑F = -2T sin θ, where θ is the angle from the horizontal.

For small angles, sin θ ≈ tan θ.

∑F = -2T tan θ

∑F = -2T (Δy / L)

(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m).  (This is derived by solving a second order differential equation.)

In this case, k = 2T/L, so the frequency is:

ω = √((2T/L) / m)

ω = √(2T / (mL))


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Answers

Answer:

1124923453 electrons

Explanation:

The formula for charge in coulomb ,C =Current in amperes, A * Time in seconds, s

Given in question ;

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Answers

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You don't need it to solve the problem.  The answer would be the same
if the blow drier were operating from a 440-V outlet, or any other kind
(just as long as the blow drier is designed to operate on whatever voltage
it was plugged into.)

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Cost per unit of energy = $ 0.1 per kWh
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===============================

The other answer has his significant figures correct, but shows what happens
when you don't carry your units all the way through.
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Answer:

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