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Andrews [41]
3 years ago
12

An ant is crawling along a straight wire, which we shall call the x axis, from A to B to C to D (which overlaps A), as shown in

the figure below. O is the origin. Suppose you take measurements and find that AB is 31 cm , BC is 12 cm , and AO is 7 cm .(Figure 1)
A. What is the ant’s position at point A?


B. What is the ant’s position at point B?


C. What is the ant’s position at point C?


D. What is the ant’s position at point D?

Physics
1 answer:
Keith_Richards [23]3 years ago
6 0

Answer:

Ants position at point A = -7 cm

Ants position at point B = 38 cm

Ants position at point C = 26 cm

Ants position at point D = -7 cm

Explanation:

In this question, we are dealing with displacement which is the change in position of an object.

Now, we are told the journey began from A to B and then from C to D.

Now, AB = 31 cm and AO = 7cm

But along AB, we have the origin O.

Since A is on the left hand side of the origin, it means it is negative. Thus, position A = -7cm

Then position B = 31 - (-7) = 31 + 7 = 38 cm

Since BC = 12cm, then position C = 38 - 12 = 26 cm

Position D is same as position A = -7cm

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Un bloque de 20kg de masa se desplaza horizontalmente en la dirección de eje X por acción de una fuerza horizontal variable F =
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a) El trabajo realizado por esta fuerza mientras el bloque se mueve desde la posición x = + 10 m hasta la posición x = + 20 m es 900 joules.

b) La rapidez del bloque en la posición x = + 20 metros es aproximadamente 5.701 metros por segundo.

Explanation:

a) El trabajo expermentado por el bloque (W), medido en joules, es definida por la siguiente ecuación integral:

W = \int\limits^{x_{max}}_ {x_{min}} F(x) \, dx (1)

Donde:

x_{min}, x_{max} - Posiciones mínima y máxima del bloque, medidos en metros.

F(x) - Fuerza horizontal aplicada al bloque, medida en newtons.

Si conocemos que F(x) = 6\cdot x, x_{min} = 10\,m y x_{max} = 20\,m, entonces el trabajo realizado por esta fuerza es:

W = \int\limits^{20\,m}_{10\,m} {6\cdot x} \, dx (2)

W = 6\int\limits^{20\,m}_{10\,m} x\, dx

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W = 3\cdot [(20\,m)^{2}-(10\,m)^{2}]

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W = K_{f}-K_{o} (3)

Donde son K_{o}, K_{f} las energías cinéticas traslacionales inicial y final, medidos en joules.

Al aplicar la definición de energía cinética traslacional, expandimos y simplificamos la ecuación como sigue:

W = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (4)

Donde:

m - Masa del bloque, medido en kilogramos.

v_{o}, v_{f} - Rapideces inicial y final del bloque, medidos en metros por segundo.

\frac{2\cdot W}{m} = v_{f}^{2}-v_{o}^{2}

v_{f} = \sqrt{\frac{2\cdot W}{m}+v_{o}^{2}}

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v_{f} = \sqrt{\frac{900\,J}{2\cdot (20\,kg)}+10\,\frac{m^{2}}{s^{2}}  }

v_{f} \approx 5.701\,\frac{m}{s}

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