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3 years ago

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A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before

striking the water

1 answer:

DENIUS [597]3 years ago

4 0

Answer:

14.05m/s

Explanation:

We are given that

Horizontal speed of diver= $v_x=1.2m/s$

Distance, y=-10 m

It is taken as negative because the diver goes downward.

There is no air resistance therefore, there is acceleration due to gravity .

We know that

$g=-9.8 m/s^2$

Initial vertical velocity, $u_y=0$

$v^2-u^2=2as$

Using the formula

$v^2_y-0=2\times (-9.8)(-10)$

$v_y=\sqrt{2(-9.8)(-10)}$

$v_y=14 m/s$

$v=\sqrt{v^2_x+v^2_y}$

Using the formula

$v=\sqrt{(1.2)^2+(14)^2}=14.05m/s$

Hence, the speed of diver before just striking the water=14.05m/s

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Si un ciclista se mueve a una velocidad de 5 m/s y acelera 1 m/s2, a los 10 segundos su velocidad será

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Answer:

A los 10 segundos su velocidad será 15 $\frac{m}{s}$

Explanation:

La aceleración de un objeto es una magnitud que indica cómo cambia la velocidad del objeto en una unidad de tiempo.

En otras palabras, la aceleración relaciona los cambios de la velocidad con el tiempo en el que se producen, es decir que mide cómo de rápidos son los cambios de velocidad:

Una aceleración grande significa que la velocidad cambia rápidamente.
Una aceleración pequeña significa que la velocidad cambia lentamente.
Una aceleración cero significa que la velocidad no cambia.

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En este caso:

a= 1 $\frac{m}{s^{2} }$

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vinicial= 5 $\frac{m}{s}$
tiempo= 10 s

Reemplazando:

$1\frac{m}{s^{2} }=\frac{vfinal - 5\frac{m}{s} }{10 s}$

Resolviendo se obtiene:

1 $\frac{m}{s^{2} }$ *10 s= vfinal - 5 $\frac{m}{s}$

10 $\frac{m}{s}$ = vfinal - 5 $\frac{m}{s}$

10 $\frac{m}{s}$ + 5 $\frac{m}{s}$ = vfinal

15 $\frac{m}{s}$ = vfinal

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A 0.001kg bullet is fired with a velocity of 800m/s into a soft wood of mass 1kg resting on a smooth surface. Find the final vel

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The final velocity of the bullet+block is 0.799 m/s

Explanation:

We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.

Mathematically, we can write:

$mu+MU=(m+M)v$

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$v=\frac{mu}{m+M}=\frac{(0.001)(800)}{0.001+1}=0.799 m/s$

Learn more about conservation of momentum:

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$V = \frac{kQr^2}{R^3}$

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