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]
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t
Case I
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When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C
Case II
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When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C
Case III
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When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C
Case IV
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When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C
Case V
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When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
C . plate a is negatively charged and plate b is positively charged
Explanation:
Given that,
(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :
(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :
Therefore, this is the required solution.
Answer:
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