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dimulka [17.4K]
4 years ago
15

A 5.0 kg box is on a frictionless surface. Two forces act on the box as shown below. 

Physics
1 answer:
lys-0071 [83]4 years ago
8 0
Hello!

So we have a 5kg box with two forces acting on it in opposite directions. So by Analysis of the diagram we can conclude that the box will be moving in the positive x direction( or to the right) with a constant acceleration of: F(net)=ma; so a=F/m=5N/5kg=1m/s^2; with a total net force of 5N

Hope this helps. Any questions please just ask and I’ll get right back to you. Thank you kindly.
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alexandr1967 [171]

Answer:

136.67 watts

Explanation:

P=W/t

P=4100/30

P= 136.67 watts

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3 years ago
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Stells [14]

Answer:

Light.

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Energy.

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Explanation:

5 0
4 years ago
One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t
Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2\frac{m}{s^{2} }

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=FΔx=(0.400m)\\

W_{ext}=20.4 Nm

U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2}  \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}

To determine, the acceleration we solve the following equation for a:

F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }

8 0
4 years ago
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Answer:

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\Delta W = m\cdot g\cdot \Delta z (1)

Donde:

m - Masa del montañero, medido en kilogramos.

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

\Delta z - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que m = 65\,kg, g = 9,807\,\frac{m}{s^{2}} y \Delta z = 500\,m, entonces el trabajo realizado por el montañero para subir es:

\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)

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7 0
3 years ago
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Blababa [14]

<em>your answer would be <u>C</u></em>

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hope this helped you- have a good day bro cya)

3 0
3 years ago
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