g A 45.0 - kg girl is standing on a 150. - kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is
2 answers:
Answer:
her velocity relative to the surface of the ice is 1.154 m/s
The positive velocity indicates that her velocity relative to the surface of the ice is in the same direction with her velocity relative to plank.
Explanation:
Given;
mass of the girl, = 45.0 - kg
mass of the plank, = 150 kg
velocity of the girl relative to the plank, = 1.5 m/s
Velocity of the plank relative to the ice, = ?
Velocity of the girl relative to the ice, = ?
Thus, velocity of the girl relative to the ice can be calculated as;
=
+
= 1.5 +
-----------equation (i)
From the principle of conservation of energy;
45 + 150
= 0
3 + 10
= 0 ------------equation (ii)
From equation (i); =
- 1.5
Substitute in in equation (ii)
3 + 10(
- 1.5) = 0
3 + 10
-15 = 0
13 = 15
= 15/13
= 1.154 m/s
The positive velocity indicates that her velocity relative to the surface of the ice is in the same direction with her velocity relative to plank.
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Answer:
24445.85 J/s
Explanation:
Area, A = 300 m^2
T = 33° C = 33 + 273 = 306 k
To = 18° C = 18 + 273 = 291 k
emissivity, e = 0.9
Use the Stefan's Boltzman law
Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.
The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4
By substituting the values
E = 24445.85 J/s