A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa
1 answer:
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Answer:
291.598 N-m
291.6 N-m
Explanation:
Let's first take a look at the free bodily diagrammatic representation.
The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).
Let's first determine our angle θ from the diagram
To find angle θ ; we have :
tan θ =
tan θ =
tan θ = 1.333
θ = tan⁻¹ (1.333)
θ = 53.13°
Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.
We have:
where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:
Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m
b) From the second diagram, taking the moment at point B ,
we have:
where Force(F) = 1350 N and θ = 53.13° ; we have:
Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m
The first law of thermodynamics says that the variation of internal energy of a system is given by:
where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is
where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is