Gravity Probe B concludes its 50-year quest

Results of an experiment conceived around 1960 to test general relativity and launched in 2004 were announced at a NASA press conference earlier this month: Albert Einstein’s theory passed. The experiment featured four (for redundancy) gyroscopes—spinning, niobium-covered spheres—orbiting 642 km above Earth . The goal was to measure the precession induced in the gyroscopes by two general relativistic effects. The easier-to-measure geodetic effect influences any spinning object orbiting a mass. The second effect, frame dragging, arises when the spacetime-distorting mass, here Earth, is itself spinning.Gravity Probe B was not the first to measure the two effects, but it was designed to measure them independent of each other and to extraordinary precision. The gyroscopes are the most perfectly spherical objects ever fabricated. They needed to be, lest the general relativistic precessions be swamped by those arising from Newtonian torques. To measure the spin of those featureless spheres, the experimenters cooled them below niobium’s superconducting transition; the superconducting metal then produces a magnetic field, parallel to its spin axis, that can be measured with a superconducting quantum interference device. In the end, the experiment was a qualified success. It measured the geodetic effect to 0.3% precision, but stray charges on the gyroscopes and their housings limited the precision of the frame-dragging measurement to 20%. In both cases other efforts have achieved comparable results.
by
Steven K. Blau

Crumple zone

The crumple zone is a structural feature mainly of automobiles. Crumple zones have also been incorporated into railcars in recent years.They are designed to absorb the energy from the impact during an accident by controlled deformation.Crumple zones work by managing crash energy, absorbing it within the outer parts of the vehicle, rather than being directly transmitted to the occupants, while also preventing intrusion into or deformation of the passenger cabin. This better protects car occupants against injury. This is achieved by controlled weakening of sacrificial outer parts of the car, while strengthening and increasing the rigidity of the inner part of the body of the car, making the passenger cabin into a 'safety cell', by using more reinforcing beams and higher strength steels. Impact energy that does reach the 'safety cell', is spread over as wide an area as possible to reduce its deformation.When a vehicle and all its contents, including passengers and luggage are travelling at speed, they have inertia which means that they will want to continue forward with that direction and speed (Newton's first law of motion). In the event of a sudden deceleration of a rigid framed vehicle due to impact, unrestrained vehicle contents will continue forwards at their previous speed due to inertia, and impact the vehicle interior, with a force equivalent to many times their normal weight due to gravity. The purpose of crumple zones is to slow down the collision and to absorb energy, to reduce the difference in speeds between the vehicle and its occupants.

Momentum

5) A body of mass 10Kg moving with a velocity 20m/s along a straight line collides with another body of mass 8Kg moving in the same direction with a velocity 5m/s. After collision if the velocity of the former is reduced to 16m/s, calculate the final velocity of the latter.

Work,Energy and Power

3) A man cycles up a hill of slope 1 in 20 with a velocity 6.4Km/h along the hill. The weight of the cycle and the cyclist is 98Kgwt.What is the horse power developed(Neglect Friction)

4) The bob of a pendulum is released from horizontal position A. If the length of the pendulum is 2.5m, what is the speed with which the bob reaches the lowermost point B, given that it dissipates 5% of its initial energy against air resistance.

Work-Energy

Question 1

A person skis down a 20 m long snow slope which makes an angle of 25° with the horizontal.

The total mass of the skier and skis is 50 kg.  There is a constant frictional force of 60 N opposing the skier's motion.  The speed of the skier as he/she descends from the top of the slope is 2,5 m·s^-1

1.1) Calculate the magnitude of the net force parallel to the slope   experienced

        by the person.        

1.2)  Calculate the maximum speed of the skier at the bottom of the 20 m slope

Question 2

A bus, of mass 8000 kg, is at point A and is moving at a speed of 20 ms^-1. As it passes point B, which is 100 m from A, its speed has decreased to 5 ms^-1.

2.1) Calculate the difference in mechanical energy between A and B.

2.2) Calculate the magnitude and direction of the average frictional force.                                                                                                           

Electric Field- answer

1.1)            The space around a charge where its effect can be felt is called electric field. An electric charge placed at any point in it experiences an electric force. Strength or intensity of the electric field at any point is defined as the force acting on a unit positive charge placed at that point. The unit of electric intensity in SI is N/C or V/m

                E=F/q

1.2)     

                                                            

                                       

1.2)            Electric field at P due to Q1:

               

                                                           E₁ = kQ/r²

                                                              = (9x 10⁹ x14x10^-6)/ 1²

                                                              = 126x10³ NC^-1   to the left

            Electric field at P due to Q2:

  

                                                         E₂ = kQ/r²

                                                              = (9x 10⁹ x20x10^-6)/ 2²

                                                              = 45x10³ NC^-1   to the left

                                                      E_net= E₁+ E₂

                                                            = 126x10³ + 45x10³ = 171 x 10³ NC^-1   to the left

2)                                                                                                                                        V=450V

d=0.15m

     Therefore electric field between the plates, E = V/d

                                                                              = 450/0.15

                                                                           E= 3000 V/m

                   Charge of electron, q = e = 1.6x 10^-19 C

              Force of the electron , F = qE

                                                     =  1.6x 10^-19x 3000 = 4.8x 10^-16 N

       Acceleration of the electron,  a= F/m

                                                        = (4.8x 10^-16)/ ( 9.1x 10^-31)

                                                        = 0.5275 x 10¹⁵ ms^-2

        Considering the motion of the electron from the negative plate to the positive plate, u=0 ,

        a=0.5275 x 10¹⁵ ms^-2 ; S=0.15 m

         S= ut+ at²/2

                      0.15=(0.5275x 10¹⁵xt²)/2

         

          t² = (2x0.15)/ (0.5275x 10¹⁵)

          t = [(2x0.15)/ (0.5275x 10¹⁵)]^1/2 = 2.385 x 10^-8 s

     The velocity with which the electron hits the positive plate, v = u + at

                 i.e,  v = 0 + ( 0.5275x 10¹⁵ x 2.385 x 10^-8 )  

                           v = 1.258 x 10⁷ ms^-1

                                                                                

Electric Field

Question 1(Grade 12 SA:Year2010)

Two point charges Q1 and Q2 a distance 4m apart are shown below .The charge on Q1 is -14μC and the charge on Q2 is +20μC.

1.1)            Define the electric field at a point in space.

1.2)            Draw the electric field pattern due to these two charges.

1.3)            Calculate the net electric field at point P situated 2m from Q2.                                                                                                                                                                                                           

Question 2

Two charged metallic plates are placed 15cm apart in vacuum and are kept at a p.d of 450V. An electron is released from rest at a point close to the negative plate .How long it will take to reach the other plate and with what velocity?(Mass of electron = 9.1x 10^-31Kg and charge of electron = 1.6x 10^-19C).