At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if <em>n</em> is the number of moles of this gas, then
<em>n</em> / (19.2 L) = (1 mole) / (22.4 L) ==> <em>n</em> = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / <em>n</em> ≈ 14.0 g/mol
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
It's hard to tell what's going on down there in the corner with the resistor and the ammeter. There seems to be as many as 3 or 4 wires in and out of the ammeter, which would be wrong. A real ammeter only has two ... one in and one out. (Same for a resistor.)
It's hard to say whether this circuit works, until we can clearly understand how everything is hooked up in that corner of the drawing.
Answer:
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Explanation:
It is given that,
Radius of the circle, 
The area of the circle is given by :
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

or
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As there is no uncertainty given in the radius of the circle. So, the area of the circle is
. Hence, this is the required solution.