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allochka39001 [22]
3 years ago
11

In one eight-hour workday, a forklift operator at a particular distribution warehouse lifts a total of 100,000 N of boxes. Each

box must be lifted to a height of 1.5 m and carried an average of 10 m to the shipping dock. What is the net work done against gravity?
Physics
1 answer:
german3 years ago
3 0

Answer:

150000 J

Explanation:

Only the work that is used to lift the boxes up by 1.5 m is counted against gravity. The work that is used to move 10m to the dock is not.

This work done to lift up by 1.5m is the product of force lifted F = 100 000 N and the vertical distance in height h = 1.5m

W = F*h = 100000*1.5 = 150000 J

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Answer:  1.  0.19

               2.  0.33

               3.  0.47

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An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth
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Note: The answer choices are :

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Explanation:

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emf = l (v \times B)

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As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. v \times B = vB sin90 = vB\\ ,

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The magnitude of the electric field potential difference between the wingtips increases

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How does the role of blood protect the body
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A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces
snow_lady [41]

Answer:

<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>

  • <u><em>Initial velocity = 33 m/s</em></u>
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  • <u><em>Average velocity = (33 + 0) / 2  m/s = 16.5 m/s</em></u>

Explanation:

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4 0
3 years ago
Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci
ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

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3 years ago
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