Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature
m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).
H = 6.9 kj = 6.9 *1000J = 6900 J
6900 = 200*0.444* (θ₂ - 22)
6900/(200*0.444) = θ₂ - 22
77.70 = θ₂ - 22
θ₂ - 22 = 77.7
θ₂ = 77.7 + 22 = 99.7
So initial temperature before cooling ≈ 100°C . Option C.
Answer:
M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b]
ΔR is the difference in density between the gas inside and surrounding the balloon.
R[b] is the density of gas inside the baloon.
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Let V be the volume of helium required.
Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V
U = 1.225gV newtons
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Weight of Helium = Volume of Helium * Density of Helium * g
W[h] = 0.18gV N
Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N -----
Weight of 260kg = 2549.7 N
Then to lift the whole thing, F > 2549.7
So minimal F would be 2549.7
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1.045gV = 2549.7
V = 248.8 m^3
Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)
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Let the density of the surroundings be R
Then U-W = (1-0.9)RgV = 0.1RgV
So 0.1RgV = 2549.7 N
V = 2549.7 / 0.1Rg
Assuming that R is again 1.255, V = 2071.7 m^3
Then mass of hot air required = 230.2 * 0.9R = 2340 kg
Notice from this that M = 2549.7/0.9Rg * 0.1R so
M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)
M[min] = M[basket] * ΔR/R[b]
Answer:I think its the first one
Explanation:
What property/properties of water allowed the water to stick to the penny?
Solution's:
Explanation:
By these properties, the water molecules gets attracted to the other molecules. Hence, this makes the water get stick to the penny.
Answer:
Completion between two organism is when they fight over resources inadvertently causing the use and destruction of recourses making it more scares. if they don't adapt they die out.
