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Paladinen [302]

3 years ago

7

Una ave vuela a una velocidad constante de 15m/s en una trayectoria rectilínea. Si dura una hora volando ¿cuanta distancia habrá

recorrido?

1 answer:

earnstyle [38]3 years ago

3 0

Answer:

54,000

Explanation:

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from rarefaction to rarefaction for a longitudinal wave

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The gravitational potential is a distance z=0.17 m away from a distribution of mass that has the following equation:

sveta [45]

gravitational potential is given as

$V = \frac{GM}{R^2}(R^2 + Z^2)^0.5$

$E = -\frac{dV}{dz}$

$E = -\frac{GM}{R^2}\frac{d}{dz}(R^2+z^2)^0.5$

$E = -\frac{GM}{R^2}*0.5(R^2+z^2)^-0.5*2z$

$E = -\frac{GM*z}{R^2*(R^2 + z^2)^0.5}$

now plug in all values given

M = 110 kg

R = 0.55 m

z = 0.17 m

$E = -\frac{6.67 * 10^{-11}* 110*0.17}{0.55^2*(0.55^2 + 0.17^2)^0.5}$

$E = 7.16 * 10^{-9} N/kg$

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3 years ago

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

HACTEHA [7]

Explanation :

Speed of car, $v_1=50\ km/h=13.8\ m/s$

Kinetic energy of the car, $KE_1=400\ J$

Speed of car, $v_2=100\ km/h=27.7\ m/s$

Let $KE_2$ is the kinetic energy of the car when it is moving with 100 km/h.

$\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}$

$\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}$

$KE_2=KE_1(\dfrac{v_2}{v_1})^2$

$KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2$

$KE_2=1611.6\ J$

Initial velocity of both cars are 0. Using third equation of motion :

So, $S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}$

and $S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}$

t is same. So,

$\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}$

$\dfrac{S_1}{S_2}=\dfrac{1}{2}$

$S_2=2\ S_1$

So, the braking distance at the faster speed is twice the braking distance at the slower speed.

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