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The linear speed of a particle at the equator of the earth due to its spin motion is V . What is its linear speed latitude 60^(@) ?
If omega_(E ) is the angular velocity of the earth rotating about its own axis and omega _(H) is the angular velocity of the hour of a clock , then
What is the linear velocity of a person at equator of the earth due to its spinning motion? (Radius of the earth =6400km )
Which is large : angular velocity of the earth about its axis or angular velocity of hour hand of a watch ?
Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth =6400 km and radius of the orbit of the earth about the sun =1.5xx10^8km.
The angular momentum of an electron due to its spin is given as ………..
The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60^@ latitude becomes zero is
The acceleration due to gravity at the poles is 10ms^(-2) and equitorial radius is 6400 km for the earth. Then the angular velocity of rotaiton of the earth about its axis so that the weight of a body at the equator reduces to 75% is
The angular velocity of earth at present is omega . What should be its angular velocity so that the body lying at the equator flies off
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