The solar energy strikes the deck at the rate of 1400 W on every square metre.
1 answer:
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Answer:
T = 273 + (-50) = 273 – 50 = 223 K
R = 188.82 J / kg K for CO2
Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg /
T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa
Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg /
Answer:
Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,
So, the surface charge density on the sphere is .
Answer:
Q=1005 J
t= 0.67 sec
Explanation:
Lets take condition of room is 1 atm and 25°C.
Heat capacity ,c = 21 J /K.mol
If we assume that air is ideal gas that
P V = n R T
V= 107250 L
At STP number of moles given as
V=22.4 L at S.T.P.
n=4787.94 moles
n= 4.784 Kmoles
So heat required to raise 10°C temperature
Q = n x c x ΔT
Q = 4.78794 x 21 x 10
Q=1004.64 J
Time t
t= Q/P
P= 1.5 KW
t = 1.004.64 /1.5
t= 0.66 sec