Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is
1 answer:
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1) the weight of an object at Earth's surface is given by 
, where m is the mass of the object and 
is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
2) On Mars, the value of the gravitational acceleration is different:
. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: 
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span><span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>
<span>
</span>
2) On Mars, the value of the gravitational acceleration is different:
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span>
</span>
</span>
Answer:
46.2 rad/s2
Explanation:
Angular acceleration works very similar to linear acceleration, it follows this equation:
Where:
γ: angular acceleration
Mt: torque
J: moment of inertia of the load from its turning axis
Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.
The wheels act like disks. For disks the moment of inertia is:
The wheel walls act like annular rings, for these the moment of inertia is:
The tread acts like a hoop, as in mass concentrated into a circunference, for these:
The axle acts like a rod, which is the same as the disk:
The drive shaft acts like a rod too:
SO, the total moment of inertia is:
J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft
J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2
Finally the angular acceleration is: