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3 years ago

Un patinador recorre un parque de 7.2 km de largo a una velocidad de 5 m/s ?cuantas horas duro su recorrido?

Physics

2 answers:

icang [17]3 years ago

5 0

T = d/v

7.2 kilómetros es lo mismo que 7200 metros.

Entonces, coloque 7200 sobre la velocidad (5m / s).

Dará 1440 segundos.

1440 segundos, que es lo mismo que 24 minutos.

Por lo tanto, la respuesta es:

En minutos: 24

En segundos: 1440

En horas: 0.4

pogonyaev3 years ago

4 0

Si velocidad es de 5m/s y esta recorriendo un parke de un tamaño de 7.2 kilometros, su recorrido le duro 1 hora y 44 minutos.

Dividi los kilómetros del parque por la velocidad, dándomela la hora que duró su recorrido

7.2÷5=1.44

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Answer:

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3 years ago

What is the physical properties of radon

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Answer:

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Explanation:

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3 years ago

A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. The d

-BARSIC- [3]

Answer:

a) $t=6.37s$

b) $t=3.3333s$

Explanation:

The knowable variables are the initial hight and initial velocity

$s_{o}=80ft$

$v_{os}=64ft/s$

The equation that describes the motion of the ball is:

$s=80+64t-16t^{2}$

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

$0=80+64t-12t^{2}$

a) Solving for t, we are going to have two answers

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

a=-16

b=64

c=80

$t=-1.045 s$ or $t=6.378s$

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

$v_{y}=v_{o}+gt$

at maximum point the velocity is 0

$0=64-32.2t$

Solving for t

$t=1.9875 s$

Now, we must know how much distance does it take to reach maximum point

$s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft$

So, the ball pass the top of the building on its way down at 160 ft

$160=80+64t-16t^{2}$

Solving for t

$t=2s$ or $t=3.333s$

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

4 0

3 years ago

Read 2 more answers

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3 years ago