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3 years ago

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The two forces in each pair can either both act on the same body or they can act on different bodies. The two forces in each pai

r can either both act on the same body or they can act on different bodies. true false

1 answer:

kotykmax [81]3 years ago

5 0

Answer: The given statement is false.

Explanation:

According to Newton's third law of motion, every action has an equal and opposite reaction. So, when we apply force in one direction on an object then the object also applies a force in the opposite direction.

Hence, it is true that two forces in each pair of forces act in opposite directions.

For example, when we push a wooden box of 20 kg in the forward direction then the box will also apply a force in the opposite direction.

But the statement two forces in each pair can either both act on the same body or they can act on different bodies is false.

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VERY URGENT GIVING BRAINLY

lakkis [162]

Answer:

A

Explanation:

There is a mechanical advantage in this system....

she only needed 100 N to lift 500 N

1:5 she lifted it 2 meters, so if there WAS NO friction, she would have to pull in 5 x 2 = 10 meters

but there IS friction so she must have pulled in MORE than 10 m

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2 years ago

Which phase of matter contains particles that split into ions and electrons?

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Answer:

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Explanation:

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3 years ago

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.

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3 years ago

At sunrise two old women started to walk towards each other. one started from point a and went towards point b while the other s

konstantin123 [22]

Let the distance between points a and b = d

For the women starting from a her velocity be $V_1$ and for the women starting from b her velocity be $V_2$

Let the rising time of sun = x AM

We have $V_1$ = $\frac{d}{12+4-x}$

and we also have $V_2$ = $\frac{d}{12+9-x}$

At they meet means, they both travel a combined distance of d

So we have $V_1*(12-x)+V_2*(12-x)=d$

Substituting velocity values we will get

$\frac{d}{12+4-x}*(12-x)+\frac{d}{12+9-x}*(12-x)=d\\ \\ \frac{12-x}{12+4-x}+\frac{12-x}{12+9-x}=1\\ \\ \frac{12-x}{16-x}+\frac{12-x}{21-x}=1\\ \\ 252-21x-12x+x^2+192-12x-16x+x^2=336-16x-21x+x^2\\ \\ x^2-24x+108=0\\ \\ (x-6)(x-18)=0\\ \\$

x= 6 AM or 18 AM, but 18 AM is not possible,

Sun rise time is 6 AM

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