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insens350 [35]
3 years ago
11

Una profesora de física que se encuentra en la tierra aplica un examen a sus estudiantes que están en una nave espacial que se d

esplaza a velocidad v = 0,6c con respecto a la tierra. En el instante en que la nave pasa frente a la profesora, ésta indica el inicio del examen. La profesora quiere que sus alumnos cuenten con dos horas (tiempo de la nave espacial) para completar el examen.
Physics
1 answer:
GuDViN [60]3 years ago
7 0

Answer:

t´ =  2,5 h

Explanation:

Dado que a velocidades "comparables" a la velocidad de la luz, los principios de la física deben ser tratados a la luz de los conceptos relativistas, es necesario aplicar el factor de Lorentz al tiempo que los estudiantes tomaran como base para la elaboración de su examen-

El factor de Lorentz  γ =  1 / √1 - (v²/c²)

En ese factor c es la velocidad de la luz y v es la velocidad a la que se mueve el móvil ( en este caso la nave espacial que va a 0,6 *c )

La otra consideración que hay que tomar en cuenta es que cuando se tienen dos sistemas inerciales con velocidades relativas comparables a la velocidad de la luz, las coordenadas espaciales ( y el tiempo que es la cuarta coordenada ) se relacionan según:

(Movimiento exclusivo en la dirección del eje x )

x´  =  ( x - vₓt )*γ      y´= y    z´ = z      t´  = ( t - vₓ/c² ) *γ

Donde las coordenadas s distinguen por ´ y sin ´ para identificar a los dos sistemas, luego entonces:

t´ ( tiempo en la nave espacial )

t´  = [ t  - (0,6*c/c² ] *γ    ⇒    t´ =  t - 0,36/c / √1 - (0.6*c)²/c²

Dado que   t = 2 h   y que la velocidad de la luz c es  300000 km/s expresada en horas es  300000*3600 = 108*10⁷,el numerador se puede aproximar a 2 horas

t´ =  2 /√ 64*10⁻²

t´ = 2 / 0,8  h

t´ =  2,5 h

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ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

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w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

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⇒n=\frac{24}{2.2}=11(approx)

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3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago
A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t
WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s

So, the required time is 2.2 seconds.

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2 years ago
Some neodymium glass lasers can provide 100TW of power in 1.0 ns pulses at a wavelength of 0.26 micrometers. how much energy is
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Answer:

 E = 10⁵ J

Explanation:

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Power, P = 100 TW

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The energy of a single pulse is:-

Energy = Power x time

 E = P t

 E = 100 x 10¹² x 1 x 10⁻⁹

 E = 10⁵ J

The energy contained in a single pulse is equal to 10⁵ J

7 0
3 years ago
2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is
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