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Kazeer [188]
2 years ago
7

A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude r

eduction of a factor of 0.96 due to a small velocity dependent frictional effect.
Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value (The amplitude after N oscillations=(initial amplitude)x(factor)^N).
Physics
1 answer:
True [87]2 years ago
5 0

The time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value

t= 14.448 s

This is further explained below.

<h3>What is an oscillator?</h3>

An electronic oscillator is a circuit that generates a periodic, oscillating electronic signal. This signal is often a sine wave, but it may also be a square wave or a triangle wave. Oscillators are devices that change the direct current that is supplied by a power source into an alternating current signal.

Generally, the equation for amplitude  is  mathematically given as

A_n = A_0(0.96)^n

equal to the half-energy amplitude

A_0(0.96)^n =\frac{ A_0}{\sqrt{2}}

lnA_0(0.96)^n = ln\frac{Ao}{\sqrt{2}}

ln(A_0)+n*ln(0.98) = ln(A_0)-ln(\sqrt{2})

n = 17.2

In conclusion, the time is

t = nT

t= 14.448 s

Read more about oscillators

brainly.com/question/17133973

#SPJ1

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A 53 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.36. The acceleration of gravity is 9.8 m
tino4ka555 [31]

Answer:

42.6 m

Explanation:

mass of crate m = 53 kg

coefficient of kinetic friction, μ = 0.36

acceleration due to gravity, g = 9.8 m/s^2

Force, F = 372.098 N

Net force, f = F - friction force

f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8

f = 185.114 N

acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2

initial velocity, u = 0

time, t = 4.94 s

s = ut + 1/2 at^2

s = 0 + 1/2 x 3.49 x 4.94 x 4.94

s = 42.6 m

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The flow control circuit design that can best control an overrunning load is the ____ circuit.
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The meter out circuit is the flow control circuit design that can most effectively control an overrunning load.

The meter-out circuit can be very accurate, but are not efficient. The meter-out circuit can control overrunning as well as opposing loads while the other one method must be used with opposing loads only. The choice of flown control valve method and the location of the flow control in the circuit are dependent on the type of application being controlled.

<h3>What is a Circuit ?</h3>

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  • The directional valve shifts, causing the actuator to move faster than pump flow can fill it due to an overrunning load. Oil is leaking from one side, whereas there is none on the other.

Hence, flow control circuit design that can best control an overrunning load is the opposing circuit

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Explanation:

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