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GuDViN [60]
3 years ago
7

En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8

y 1.4 segundos de haberse iniciado el movimiento
Physics
1 answer:
Ivanshal [37]3 years ago
6 0

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

A=10cm\\T=2s

the equation of the motion is

x=Acos(\omega t)\\

we can calculate w by using

\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi

Hence, we have that

x=10cm*cos(\pi t)\\

the speed will be

v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}

hope this helps!

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A pendulum has 895 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
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Newton's law of conservation states that energy of an isolated system  remains a constant. It can neither be created nor destroyed but can be transformed  from one form to the other.

Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

Where m is the mass of the pendulum.

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h is the height from the bottom z the ground.

At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

Where m is the mass of the pendulum

v is the velocity of the pendulum

At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.

Also as it has been mentioned energy can neither be created nor destroyed hence the entire potential energy is converted to kinetic energy at the bottom and would be equivalent to 895 J.

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In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

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F=ma
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Hope this helps!
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The force of gravity produces acceleration in all C. freely falling objects and this is known as acceleration due to gravity

Explanation:

A body is said to be in free fall when there is only one force acting on the body: the force of gravity.

Gravity is a force that acts downward, i.e. towards the Earth's centre.

If we are near the Earth's surface, the magnitude of the force of gravity on a body is given by

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Solving for the acceleration,

a=\frac{F}{m}

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Therefore, every object in free-fall accelerates at 9.8 m/s^2 towards the ground.

Learn more about free fall here:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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