Two piston hydraulic problems
1 answer:
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Answer:
V = 42.41cm^3
Explanation:
In order to calculate the volume of the solid, you use the following formula:
where
r: radius of the circular base of the cone = 3 cm
h: height from the circular base to the peak of the cone = 4.5 cm
You replace the values of r and h in the formula for the volume V:
hence, the volume of the solid is 42.41 cm^3
Answer:
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
Explanation:
For this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where q is the charge inside the surface.
In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product
∫ E dA = q_{int}/ ε₀
The area of a sphere is
A = 4π r²
- The electric field for a distance r < R_int
The charge inside is zero, so the electric field
E = 0 r <R_in t
- The field for a radio inside the shell
Let's use the concept of density
ρ = Q / V
q = ρ (4/3 π r³)
dq = ρ 4π r² dr
We substitute in the Gaussian equation
E ∫ dA = ρ 4π r² dr / ε₀
E 4π r² = ρ 4π/ε₀ r³ / 3
E = ρ / 3ε₀ r
We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E
E- 0 = ρ / 3ε₀ (r –R_int)
Density is
ρ = q / 4/3 π (R_out³ - R_int³)
Where R <r
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
The first law of thermodynamics states that:
where
is the variation of internal energy of the system
Q is the heat absorbed by the system
L is the work done by the system on the surrounding.
In this problem, the system absorbs 15 J of heat, so Q=+15 J (with positive sign, since it is heat absorbed by the system) while the work done by the system is L=+7 J (with positive sign, since it is work done by the system), so the variation of internal energy is
where
Q is the heat absorbed by the system
L is the work done by the system on the surrounding.
In this problem, the system absorbs 15 J of heat, so Q=+15 J (with positive sign, since it is heat absorbed by the system) while the work done by the system is L=+7 J (with positive sign, since it is work done by the system), so the variation of internal energy is
"u" stands for "units" since the unit for mass in this problem is unknown.