The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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<span>It's another energy balance equation, though: energy to start with is the same as energy that you end with. Suppose that we start a distance r0 from the Earth and end a distance r1 from the Moon, then the energy balance gives:
1 v02 - G M / r0 - G m / (D - r0) = 1 v12 - G M / (D - r1) - G m / r1
...where m is the moon's mass.
One simple limit takes D ? ? and 1 v02 ? G M / r0 (the escape velocity equation), to yield:
1 v12 ? G M / r1
v1 ? ?( 2 G M / r1 ) = 2377 m/s.</span>
Answer:
= - 26.31 kJ
Explanation:
we know that number of moles is calculated as
= 0.00476 mol
Heat absorbed by calorimeter
= 26.87 kJ
Enthalpy of combustion
= - 55290.12 kJ/mol
Negative sign shows that the heat is released
The balanced reaction
ΔHc = ΔU + Δng (RT)
-55290.12 = ΔU + (12 - 12) *(RT)
= - 26.31 kJ
Answer:
Symptoms of carbon dioxide toxicity include high blood pressure, flushed skin, headache and twitching muscles
Hope this helps :)