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3 years ago

15

Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho

rizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?

1 answer:

Yuri [45]3 years ago

3 0

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

$F_{A} + F_{B} + F_{C} = 0$ (1)

Si sabemos que $F_{A} = -3$ y $F_{B} = 5$ , entonces el valor de la fuerza que debe ejercer la persona C debe ser:

$F_{C} = -F_{A}-F_{B}$

$F_{C} = -(-3)-5$

$F_{C} = -2$

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

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Answer:

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3 years ago

Q11:A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5

natulia [17]

Answer:

v_s = 34.269 m / s

Explanation:

This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.

f ’= f $\frac{v}{v \mp v_s }$

where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer

In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together

5500 = f ( $\frac{v}{v - v_s }$ )

4500 = f ( $\frac{v}{v + vs}$ )

let's solve these two equations

$\frac{5500}{4500} = \frac{v+v_s}{v-v_s}$

1.222 (v-v_s) = v + v_s

v_s (1+ 1.22) = v (1.222 -1)

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the speed of sound in air is v = 343 m / s

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3 years ago

PLEASE HELP, THANK YOU!.. :)

Firlakuza [10]

Answer:

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<u><em>Second Reaction:</em></u>

$^{230} Th$ => $^{226} Ra + ^{4}He$

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3 years ago

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stich3 [128]

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3 years ago

Read 2 more answers

A 185 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring

denis-greek [22]

Answer:

d = 5.10 m

Explanation:

As we know that here on the plane of the inclined there is no frictional force

So in these cases we can say that total mechanical energy will always remains conserved

so here we can say that

spring potential energy = gravitational potential energy of the block

as we know from the formula

$\frac{1}{2}kx^2 = mgh$

now plug in the values in it

$\frac{1}{2}(1.60 \times 10^3)(0.10)^2 = (0.185)(9.81)h$

$8 = 1.81 h$

$h = 4.42 m$

now as we know that the angle of inclination is 60 degree and height raised is 4.42 m

so here maximum distance moved along the inclined plane will be

$\frac{h}{d} = sin60$

$d = \frac{h}{sin60}$

$d = \frac{4.42}{sin60} = 5.10 m$

6 0

3 years ago