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3 years ago

Simple Harmonic Motion

Physics

1 answer:

Delvig [45]3 years ago

7 0

- The net force is greatest at the position of maximum displacement

- The net force is zero when at the equilibrium position

Explanation:

The motion of a spring is a Simple Harmonic Motion, in which the displacement of the end of the spring is given by a periodic function of the form

$x=Asin (\omega t)$

where A is the amplitude (the maximum displacement), and $\omega$ the angular frequency of the motion.

We can analyze the net force acting on the spring by looking at Hooke's law:

$F=kx$

where

F is the net force

k is the spring constant

x is the displacement

From the equation, we notice immediately that:

The net force is the greatest when the displacement x is the greates, so at the position in which the spring has maximum compression or stretching
The net force is zero when the displacement x is zero, so when the spring crosses the equilibrium position

Learn more about forces:

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Air enters the compressor of a gas-turbine engine at 300 K and 100 kPa, where it is compressed to 700 kPa and 580 K.Heat is tran

Natasha2012 [34]

Answer:

- the fraction of the turbine work output used to drive the compressor is 50%

- the thermal efficiency is 29.6%

Explanation:

given information:

$T_{1}$ = 300 k

$P_{1}$ = 700 kPa

$T_{2}$ = 580 K

Qin = 950 kj/kg

efficiency, η = 86% = 0.86

Assume, Ideal gas specific heat capacities of air, $C_{p}$ = 1.005

k = 1.4

(a) the fraction of the turbine work output used to drive the compressor

Qin = $C_{p}$ ( $T_{3}$ - $T_{2}$ )

$T_{3}$ - $T_{2}$ = Qin / $C_{p}$

$T_{3}$ = Qin / $C_{p}$ + $T_{2}$

thus,

$T_{3}$ = $\frac{950 kJ/kg}{1.005 kj/kgK} + 580 K
[tex]T_{3}$ = 1525 K

$\frac{T_{4} }{T_{3} }$ = $(\frac{P_{4} }{P_{3} }) ^{k-1/k}$ \

$\frac{P_{4} }{P_{3} }$ = $\frac{P_{1} }{P_{2} }$

so,

$\frac{T_{4} }{T_{3} }$ = $(\frac{P_{1} }{P_{2} }) ^{k-1/k}$ \

$T_{4}$ = $T_{3}$ $(\frac{P_{1} }{P_{2} }) ^{k-1/k}$ \

= 1525 K $(\frac{100}{700} )^{1.4-1/1.4}$

= 875 K

$W_{in}$ = $C_{p}$ ( $T_{2}$ - $T_{1}$ )

= 1.005 kj/kgK (580 K - 300 K)

= 281.4 kJ/kg

$W_{out}$ = η $C_{p}$ ( $T_{3}$ - $T_{4}$ )

= 0.86 (1.005 kj/kgK) (1525 k - 875 K)

= 562.2 kJ/kg

therefore,

the fraction of turbine = $\frac{W_{in} }{W_{out} }$

= $\frac{281.4}{562.2}$

= 50%

(b) the thermal efficiency

η = ΔW/ $Q_{in}$

= ( $W_{out}$ - $W_{in}$ )/ $Q_{in}$

= $\frac{562.2}{281.4}$ /950

= 29.6%

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3 years ago

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If the seafloor is spreading why isn't earth getting bigger?

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3 years ago

a ball is thrown down at 25 m/s from a 500m tall building. how fast is it traveling when it hits the ground?

Mrrafil [7]

Answer:

The speed of the ball when it hits the ground is 102.1 m/s

Explanation:

Given;

initial velocity of ball, u = 25 m/s

distance traveled by the ball = height of the building = h = 500 m

when the ball hits the ground, the final velocity, v = ?

The final velocity of the ball is given by;

v² = u² + 2gh

where;

g is acceleration due to gravity = 9.8 m/s²

v² = (25)² + 2(9.8)(500)

v² = 10425

v = √10425

v = 102.1 m/s

Therefore, the speed of the ball when it hits the ground is 102.1 m/s

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3 years ago

A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu

Ghella [55]

The angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

<h3>What is angular frequency?</h3>

Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.

<h3>What is cyclotron?</h3>

The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.

<h3>Calculation of angular frequency:</h3>

Given,

B = 0.47 T

r = 0.68

mass of proton = 1.6x $10^{-27}$