A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
1 answer:
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Answer:
The percentage of the weight supported by the front wheel is A= 19.82 %
Explanation:
From the question we are told that
The center of gravity of the plane to its nose is
The distance of the front wheel of the plane to its nose is
The distance of the main wheel of the plane to its nose is
At equilibrium the Torque about the nose of the airplane is mathematically represented as
Where m is the mass of the airplane
is the weight of the airplane supported by the main wheel
So
Substituting values
Now the weight supported at the frontal wheel is mathematically evaluated as
Substituting values
Now the weight of the airplane is = mg
Thus percentage of this weight supported by the front wheel is 19.82 %
Answer:
x = 45 cm
Explanation:
Given that,
The length of a rod, L = 50 cm
Mass, m₁ = 0.2 kg
It is at 40cm from the left end of the rod.
We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.
The centre of mass of the rod is at 25 cm.
Taking moments of both masses such that,
The distance from the left end is 40+5 = 45 cm.
Hence, at a distance of 45 cm from the left end it will balance the rod.