The ideal gas constant is a proportionality constant that is added to the ideal gas law to account for pressure (P), volume (V), moles of gas (n), and temperature (T) (R). R, the global gas constant, is 8.314 J/K-1 mol-1.
According to the Ideal Gas Law, a gas's pressure, volume, and temperature may all be compared based on its density or mole value.
The Ideal Gas Law has two fundamental formulas.
PV = nRT, PM = dRT.
P = Atmospheric Pressure
V = Liters of Volume
n = Present Gas Mole Number
R = 0.0821atmLmoL K, the Ideal Gas Law Constant.
T = Kelvin-degree temperature
M stands for Molar Mass of the Gas in grams Mol d for Gas Density in gL.
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Answer:
F2 = 834 N
Explanation:
We are given the following for the bicycle;
Diameter; d1 = 63 cm = 0.63 m
Mass; m = 1.75 kg
Resistive force; F1 = 121 N
For the sprocket, we are given;
Diameter; d2 = 8.96 cm = 0.0896 m
Radius; r2 = 0.0896/2 = 0.0448 m
Radial acceleration; α = 4.4 rad/s²
Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²
Where r1 = (d1)/2 = 0.63/2
r1 = 0.315 m
Thus, I = 1.75 × 0.315²
I = 0.1736 Kg.m²
The torque is given by the relation;
I•α = F1•r1 - F2•r2
Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².
Thus;
0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)
>> 0.76384 = 38.115 - (0.0448F2)
>> 0.0448F2 = 38.115 - 0.76384
>> F2 = (38.115 - 0.76384)/0.0448
>> F2 = 833.73 N
Approximately; F2 = 834 N
the equation of the tangent line must be passed on a point A (a,b) and
perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle,
its center is C(0,0). And we assume that the tangent line passes to the point
A(2.3).
</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.
<span>Let's find the equation of the line parallel to the radius.</span>
<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>
det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.
let's find the equation of the line perpendicular to this previous line.
let M a point which lies on the line. so MA.AC=0 (scalar product),
it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent
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