Answer:
Explanation:
x in (-oo:+oo)
2 < (1/2)*x-3 // - (1/2)*x-3
2-((1/2)*x)+3 < 0
(-1/2)*x+2+3 < 0
5-1/2*x < 0 // - 5
-1/2*x < -5 // : -1/2
x > -5/(-1/2)
x > 10
x in (10:+oo)
(10:+oo)
10,000J will raise the temperature of 1 liter of water by 2.38 °C
Using Q = mcΔT where Q = quantity of heat supplied = 10000 J,
m = mass of 1 liter of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 1 liter = 1 dm³ = 1 × 10⁻³ m³.
So, m = ρV = 1000 kg/m³ × 1 × 10⁻³ m³ = 1 kg,
c = specific heat capacity of water = 4200 J/kg°C and ΔT = temperature change.
Making ΔT subject of the formula, we have
ΔT = Q/mc
Substituting the values of the variables into the equation, we have
ΔT = Q/mc
ΔT = 10000 J/(1 kg × 4200 J/kg°C)
ΔT = 10000 J/(4200 J/°C)
ΔT = 2.38 °C
So, 10,000J will raise the temperature of 1 liter of water by 2.38 °C
Learn more about temperature change here:
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When Salt is dissolved into a water solvent; Yes, it can be extracted through evaporation. It retains its original properties: Choice C.
The salt when dissolved in water solvent forms a solution.
They can be separated by boiling the mixture.
The solution formed is a mixture which can be separated by evaporation of water molecules do that the salt is left in the heating chamber.
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F2 equals 10 N and F4 equals 2 N
The answer is B. F=ma, so plug that into the equation (10.8=0.6a) and then solve algebraically, resulting in 18.