A 98 N ball is suspended from a cable so that it hangs 3.5 m above the earth. Find the mass of the ball and the
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Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
= 1 m³
= 10°C = 283 K
= 350 kPa
= 3 kg
= 35°C = 308 K
= 150 kPa
Now, lets apply the ideal gas equation;
=
R
=
R
/
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
= ( 3 × 0.287 × 308) / 150
= 265.188 / 150
= 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, =
/ R
= (350 × 1)/(0.287 × 283) = 350 / 81.221
= 4.309 kg
Total mass, =
+
= 4.309 + 3 = 7.309 kg
Total volume =
+
= 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
=
R
/
given that; final temperature = 20°C = 293 K
we substitute
= ( 7.309 × 0.287 × 293) / 2.77
= 614.6211119 / 2.77
= 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
let the distance of pillar is "r" from one end of the slab
So here net torque must be balance with respect to pillar to be in balanced state
So here we will have
here we know that
mg = 19600 N
Mg = 400,000 N
L = 20 m
from above equation we have
so pillar is at distance 10.098 m from one end of the slab