A piston raises a weight, then lowers it again to its original height. If the process the system follows as the piston rises is
1 answer:
Answer:
Negative
Explanation:
First law of thermodynamic also known as the law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
The first law relates relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.
The first law is mathematically given as ΔU = -
= Q - W
Where Q = Quantity of heat
W = Work done
From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.
Analyzing the pistol when it raises in isothermal and when it falls in isobaric state.The following can be said:
In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.
In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.
In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.
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To solve this problem, we use the Law of Universal Gravitation:
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is
Explanation:
From the question we are told that
The mass of the steel ball is
The length of arm is
The mass of the arm is
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as
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Generally the magnitude of torque about the athlete shoulder is mathematically represented as
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