T = ??°K, V = 15.50L, P = 870torr
PV = nRT, with P in atm, 760torr/atm
T = PV / nR = 1atm/760torr × 870torr×15.5 ÷ .0821 = 216.12 °K
Answer:
Design 2
Explanation:
I had this same question
My answer:
"Design 2 is the well-designed one, because the air molecules are the most compact and could protect the individual better than 1,3,and no air bag."
Answer:
R = 0.237 m
Explanation:
To realize this problem we must calculate the moment of inertia of the wheel formed by a thin circular ring plus the two bars with an axis that passes through its center.
The moments of inertia of the bodies are additive quantities whereby we can add the mounts inertia of the ring and the two bars.
Moment of inertia ring I1 = MR²
Moment of inertia bar I2 = 1/12 ML²
Moment of inertia disk I3 = ½ mR²
Let's calculate the moment of inertia of the wheel
I = I1 + 2 I2
I = MR² + 2 1/12 ML²
The length of the bar is ring diameter
L = 2R
I = 5.65 0.156² + 1/6 9.95 (2 0.156)²
I = 0.1375 + 0.1614
I = 0.2989 kg m²
This is the same moment of inertia of the solid disk,
Disk
I3 = I
I3 = ½ MR²
They give us disk density
ρ = M / V
M = ρ V
M = ρ (pi R² e)
Done is the thickness of the disc, in general it is e= 1 cm = 0.01 m
Let's replace
I3 = ½ ( ρ π R²) R²
I3 = ½ ρ π e R⁴
R⁴ = 2 I3 / ( ρ π e)
R = ( 2 I3 / ( ρ π e)
R⁴ = 2 0.2989 / (5990 π 0.01)
R = 0.237 m
Answer:
Ratio is 100:1
Explanation:
From hydraulic press formulas;
(F1/A1) = (F2/A2)
Applying this to the question, we can say that Force on slave cylinder (Fs) = F1 while Force on master cylinder (Fm) = F2 and their respective Areas as As and Am.
Now plugging it into the equation to obtain;
Fs/As = Fm/Am
Now, the question says the hydraulic system is designed to exert a force 100 times as big as the one put into it, thus;
Fs = 100 Fm
And so from the equation earlier, rearranging, we have,
Fs/Fm = As/Am. So, 100Fm/Fm = As/Am
So, Fm will cancel out and we have ;
As/Am = 100/1
Thus, the ratio of the area of the slave cylinder to the master cylinder is 100:1