A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t
1 answer:
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<h2>Answer: its temperature must increase.</h2>
Explanation:
In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.
In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure is neutralized.
Now, according to the First law of Thermodynamics that establishes the conservation of energy:
(1)
Where:
is the internal energy
is the heat transferred
is the work
Now, for an isobaric process:
(2)
Where:
is the pressure (<u>always positive</u>)
is the volume variation of the gas
<u />
<u>Here we have two possible results:</u>
-If the gas expands (positive ), the work is positive.
-If the gas compresses (negative ), the work is negative.
In this case we are talking about the first result (work is positive).
Then, according to the above, equation (1) can be written as follows:
(3)
Clearing :
(4)
Then, for an ideal gas in an isobaric process, part of the heat () added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.