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Jobisdone [24]
3 years ago
8

Heidi (39 kg) is walking her tiny chihuahua, Chaxi (5.60 kg), on the sidewalk. To encourage Chaxi along, Heidi pulls forward wit

h a force of 9.55 N. Identify the correct reaction force in response to Heidi’s action force.
a. The friction is less than 660 N since the beam can be moved at a constant velocity.
b. There is no friction acting on the beam since it is accelerating.
c. The friction is equal to 660 N since the beam is not accelerating.
d. The friction is greater than 660 N since the beam is not in equilibrium.
Physics
1 answer:
Ivan3 years ago
4 0

Answer:

The correct reaction force in response to Heidi's action force is:

c. The friction is equal to 660 N since the beam is not accelerating.

Explanation:

Heidi's action force does not affect the beam.  Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect.  The reaction force is what makes the dog to move because it acts on it.  According to Newton's Third Law of Motion, forces always come in action-reaction pairs.  This Third Law states that for every action force, there is an equal and opposite reaction force.  This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."

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Alika [10]

Answer:

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Explanation:

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In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

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\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

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\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

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\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

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Efficiency = 52%

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It might me same volume

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