We can solve the problem by using the first law of thermodynamics, which states that:
where
is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system
In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
Answer:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
Explanation:
The graph shown in the figure is a velocity-time graph, which means that:
- On the x-axis, the time is plotted
- On the y-axis, the velocity is plotted
Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
From the graph, it would be possible to infer additional information. In particular:
- The area under the graph represents the total distance covered by the object
- The slope of the graph represents the acceleration of the object
Answer:
22.17 degree
Explanation:
n = 1.52
Angle of incidence, i = 35 degree
Let the angle of refraction is r.
use the Snell's law
n = Sin i / Sin r
Sin r = Sin i / n = Sin 35 / 1 .52
Sin r = 0.37735
r = 22.17 degree
Thus, the ray is refracted at an angle of 22.17 degree.
Let:
Vx = the pulling component of force
Vy = the lifting component of force
Vy:
Sin(n°) = Vy/hypotenuse
hypotenuse * Sin(n°) = Vy
100N*sin(30°) = Vy
50N = Vy
Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
about 86.6N = Vx
