A body is suspended on a spring balance in a ship sailing along the equator with a speed `v`. Show that the scale reading will be very close to `W_(0)(1 pm (2 omega v)/(g))`, where `omega` is the angular speed of the earth and `W_(0)` is the reading of spring balance when the ship is at rest. Explain the plus or minus sign

A body suspended on a spring balance in a ship weighs W_(0) when the ship is at rest. When the ship begins to move along the equator with a speed v , show that the scale reading is very close to W_(0) (1+-2omegaV//g ), where omega is the angular speed of the earth.

A body is suspended from a spring balance kept in a satellite. The reading of the balance is W_1 when the satellite goes in an orbit of radius R and is W_2 when it goes in an orbit of radius 2R.

Find the average speed of a particle whose velocity is given by V=V_(0) sin omega t t where T=2pi//omega is the time of complete cycle

A vessel completely filled wit water is suspended with help of strings from a spring balance as shwon in figure. A cork is plugged at the bottom of the vessel. If the initial reading of the spring balannce is W_(0) . Then the reading of the spring balance just after the cork is removed and the liquid begins to flow out will be [rho= density of liquid & A=coss -sectional area of narrow pipe]

A pendulum having a bob of mas m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T_0 . a. Find the speed of the ship due to rotation of the earth about its axis. b. find the difference between T_0 and the earth's attraction on the bob. c. If the ship sails at speed v, what is the tension in the string ? Angular speed of earth's rotation is omega and radius of the earth is R.

A satellite moves estwards very near the surface of the Earth in equitorial plane with speed (v_(0)) . Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If R =radius of the Earth and omega be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period as observed on the earth.

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this spring, when displaced and released, oscillates with a period of 0.6s. What is the weight of the body? When 50 kg is put, the extension of spring is 20 cm.

Weight of a body depends directly upon acceleration due to gravity g . Value of g depends upon many factors. It depends upon the shape of earth, rotation earth etc. Weight of a body at a pole is more then that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda as per relation given below : g_(rot)=g-Romega^(2)cos^(2)lambda where R is radius of earth and omega is angular velocity of earth. A body of mass m weighs W_(r ) in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It observed that weight W_(m) of the same body in the moving train is different from W_(r ) . Let v_(e ) be the velocity of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the relative between earth and trainwill affect the weight of the body. Weight W_(m) of the body can be given as

Weight of a body depends directly upon acceleration due to gravity g . Value of g depends upon many factors. It depends upon the shape of earth, rotation earth etc. Weight of a body at a pole is more then that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda as per relation given below : g_(rot)=g-Romega^(2)cos^(2)lambda where R is radius of earth and omega is angular velocity of earth. A body of mass m weighs W_(r ) in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It observed that weight W_(m) of the same body in the moving train is different from W_(r ) . Let v_(e ) be the velocity of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the relative between earth and trainwill affect the weight of the body. Difference between Weight W_(r ) and the gravitational attraction on the body can be given as

A rocket starts vertically upward with speed v_(0) . Show that its speed v at height h is given by v_(0)^(2)-v^(2)=(2hg)/(1+h/R) where R is the radius of the earth and g is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed 0.9 times the escape velocity.