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mestny [16]
3 years ago
5

Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice

is pushing an identical box across the floor at a constant speed of 3.4 m/s, applying a horizontal force. a. What is the magnitude of the force that Alice is applying to the box? F = N b. With the two boxes starting from rest, explain qualitatively what Alice and Bob did to get their boxes moving at different constant speeds. a. In order to keep the box moving twice as fast, Alice had to apply a constant force that was twice as large as the force that Bob applied.b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.
Physics
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

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D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

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t = time taken to meet = 4 h

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v₁ t + v₂ t = D

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Answer:

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(b) \frac{E_{d} }{m} = 22.12 Btu/lb

Explanation:

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At the temperature (T_{1})of 800 ºR and pressure (P_{1}) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{1}) = 191.81 BTu/lb

Specific entropy (s_{1}) = 0.6956 Btu/(lb.ºR)

At the temperature (T_{2})of 600 ºR and pressure (P_{2}) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{2}) = 143.47 BTu/lb

Specific entropy (s_{2}) = 0.6261 Btu/(lb.ºR)

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(b) To calculate the exergy destruction, we will use the equation for exergy rate:

\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]

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