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3 years ago

5

The strength of the gravitational attraction depends on the mass of the objects involved and on the distance between them. The g

ravitational force between two objects will be lowest in which of the following situations?

2 answers:

astra-53 [7]3 years ago

4 0

Answer:

n just copy and paste the explanation

Explanation:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

wlad13 [49]3 years ago

3 0

Answer:

The gravitational force between two objects depends only on their masses.

i think sorry if its wrong!!

Explanation:

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Un bombero alejado d = 31.0 m de un edificio en llamas dirige un chorro de agua desde una manguera contra incendios a nivel del

Mila [183]

Answer:

El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.

Explanation:

El chorro de agua exhibe un movimiento parabólico, dado que este tiene una inclinación inicial y la única aceleración es debida a la gravitación terrestre. Las ecuaciones cinemáticas que modelan el fenómeno son:

Distancia horizontal (en metros)

$x = x_{o} + v_{o}\cdot t \cdot \cos \theta$

Donde:

$x_{o}$ - Posición horizontal inicial, medida en metros.

$t$ - Tiempo, medido en segundos.

$v_{o}$ - Velocidad inicial, medida en metros por segundo.

$\theta$ - Angulo de inclinación del chorro de agua, medido en grados sexagesimales.

Distancia vertical (en metros)

$y = y_{o} + v_{o}\cdot t \cdot \sin \theta + \frac{1}{2}\cdot g \cdot t^{2}$

Donde:

$y_{o}$ - Posición vertical inicial, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

Partiendo de la primera ecuación, se despeja el tiempo:

$t = \frac{x - x_{o}}{v_{o}\cdot \cos \theta}$

Si $x = 31\,m$ , $x_{o} = 0\,m$ , $v_{o} = 40\,\frac{m}{s}$ and $\theta = 33^{\circ}$ , entonces:

$t = \frac{31\,m-0\,m}{\left(40\,\frac{m}{s} \right)\cdot \cos 33^{\circ}}$

$t = 0.924\,s$

La altura máxima se calcula por sustitución directa de términos en la segunda ecuación. Si $y_{o} = 0\,m$ , $v_{o} = 40\,\frac{m}{s}$ , $t = 0.924\,s$ y $g = -9.807\,\frac{m}{s^{2}}$ , entonces:

$y = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (0.924\,s)\cdot \sin 33^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.924\,s)^{2}$

$y = 15.943\,m$

El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.

3 0

3 years ago

A liquid with a density of 900kg/m 3 is stored in a pressurized, closed storage tank. The tank is cylindrical with a 10m diamete

Orlov [11]

Answer:

18.62 m/s

Explanation:

Given that:

A liquid with a density of 900 kg/m 3 is stored in a pressurized, closed storage tank.

Diameter of the tank = 10 m

The absolute pressure in the tank above the liquid is 200 kPa = 200, 000 Pa

At pressure of 200 kPa ; the final velocity = 0

Atmospheric pressure at 5cm = 101325 Pa

We are to calculate the initial velocity of a fluid jet when a 5cm diameter orifice is opened at point A?

By using Bernoulli's theorem between the shaded portion in the diagram;

we have:

$Pa \ + \ \frac{1}{2} \ \delta \ v^2_1 \ + \ \delta gy_1 = P + \delta gy_2$

$\frac{1}{2} \ \delta \ v^2_1 \ = P + \delta gy_2 - \ \delta gy_1 - Pa$

$\frac{1}{2} \ \delta \ v^2_1 \ = \delta g(y_2 -y_1 )+ ( P - Pa )$

$v_1 \ = \sqrt{ \frac {2 \ \delta g(y_2 -y_1 )+ 2 ( P - Pa )}{\delta}}$

where;

Pa = atmospheric pressure = 101325 Pa

$\delta$ = density of liquid = 900 kg/m³

$v_1$ = initial velocity = ???

g = 9.8 m/s²

$y_1$ = height of the hole from the buttom

$y_2$ = height of the liquid surface from the button

$v_1 \ = \sqrt{ \frac {2*900*9.8(7 - 0.5 )+ 2 ( 200,000 - 101325 )}{900}}$

$v_1 = 18.62 \ m/s$

Thus, the initial velocity of the fluid jet = 18.62 m/s

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3 years ago

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Schach [20]

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4 years ago

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