A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.
1 answer:
Answer:
a)
b) Efficiency=76.77%
Explanation:
a)
In order to solve this problem, we can use the following formula:
the problem provides us with all the necessary information so we can directly use the formula:
b) In order to find the efficiency, we can use the following formula:
so we get:
Efficiency=76.77%
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Answer:
Explanation:
<u>Net Force And Acceleration </u>
The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by
The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N
The components of the first force are
The components of the second force are
The net force is
The magnitude of the net force is
The acceleration has a magnitude of
The direction of the acceleration is the same as the net force:
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."
Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as
Where
and the position vector
using the determinant method to expand the cross product in order to determine the torque we have
by expanding we arrive at
since we have determine the vector value of the toque, we now compare with the torque value given in the question
if we directly compare the j coordinate we have