Answer:
D. Energy is converted between kinetic and gravitational potential energy.
Explanation:
Answer:
1. Yes, it can occur adiabatically.
2. The work required is: 86.4kJ
Explanation:
1. The internal energy of a gas is just function of its temperature, and the temperature changes between the states, so, the internal energy must change, but how could it be possible without heat transfer? This process may occur adiabatically due to the energy balance:
This balance tell us that the internal energy changes may occur due to work that, in this case, si done over the system.
2. An internal energy change of a gas may be calculated as:
Assuming 
constant,
Answer:
electric field E = (1 /3 e₀) ρ r
Explanation:
For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.
The charge within our surface is
ρ = Q / V
Q ’= ρ V
'
The volume of the sphere is V = 4/3 π r³
Q ’= ρ 4/3 π r³
The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area
I E da = Q ’/ ε₀
E A = E 4 πi r² = Q ’/ ε₀
E = (1/4 π ε₀) Q ’/ r²
Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant
R = Q ’/ V’ = Q / V
How you want the solution depending on the density (ρ) and the inner radius (r)
Q ’= R V’
Q ’= ρ 4/3 π r³
E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³
E = (1 /3 e₀) ρ r
According to the following formula, the answer is 2,300 g or 2.3 kg:
Volume (m)/Mass (m) Equals Density (p) (V)
Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:
p=m/V
1.15 g/mL is equal to x g/2.00 L or x g/2,000 mL.
2,000 mL of x g = 1.15 g of g/mL
2.3 kg or 2,300 g for x g.
<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>
Its active ingredient is glucose. This medication includes 50 g of glucose per 1000 ml (equivalent to 55 g glucose monohydrate). 50 mg of glucose is present in 1 ml (equivalent to 55 mg glucose monohydrate). A transparent, nearly colourless solution of glucose in water is what is used in glucose intravenous infusion (BP) at 5% weight-to-volume.
Patients who are dehydrated or who have low blood sugar levels get glucose intravenously. Other medications may be diluted with glucose intravenous infusion before being injected into the body. Other diseases and disorders not covered above may also be treated with it.
learn more about glucose intravenous infusion refer
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