- force velocity mass time The two things that determine an object's momentum are square and square
- A 1.00-kg red cart is moving rightward with a speed of 60cm/s when it collides with a 0.50 -kg blue cart that is initially at rest.After the collision, the blue cart moving rightward with a speed of 80cm/s The red cart is still moving rightward but has slowed down to a speed of 20cm/s Enter the momentum values of each individual cart and of the system of two carts before and after the collision. Also indicate the change in momentum of the carts and of the system
- EXAM APLE 6.6.2 | The altitude of a near circular, low earth orbit (LEO) satellite is about 200 miles. (a) Calculate the period of this satellite. Solution: For circular orbits, we have (GM_(E)m)/(R^2)=m(v^2)/(R)=m(4pi ^2R^2/z^2)/(R) Solving for t tau ^2=(4pi ^2)/(GM_(E))R^3 6.6 Kepler's Third Law: The Harmonic Law 241 which-no surprise-is Kepler's third law for objects in orbit about the Earth. Let R=R_(E)+h where his the altitude of th e satellite above the Earth's surface. Then x^2=(4pi ^2)/(GM_(E))R_(E)^3(1+(h)/(R_(E)))^3 But GM_(E)/R_(E)^2=g so we have z=2pi sqrt ((R_(E))/(g))(1+(h)/(R_(E)))^3/2approx 2pi sqrt ((R_(E))/(g))(1+(3h)/(2R_(E))) Putting in numbers R_(E)=6371km,h=322km we get tau approx 90.8minapprox 1.51 hr. There is another way to do this if you realize that Kepler'; third law, being a deriv- ative of Newton's laws of motion and his law of gravitation , applies to any set of bodies in orbit about another. The Moon orbits the Earth once every 27.3 days at a radius 60.3R_(E) . Thus, scaling Kepler's third law to these values (1 month=27.3 days a nd 1 lunar unit (LU)=60.3R_(E)) we have tau ^2(months)=R^3(LU) Thus, for our LEO satellite R=(6693)/(6371)R_(E)=1.051R_(E)=(1.051R_(E))/(60.3R_(E)IL)=0.01743LU t(mondhs)=R^2a[LU]=(0.01743)^122months=0.002801months=1.51ln (b) A geosynchronous satellite orbits the Earth in its equatorial plane with a period of 24 hr.Thus,it seems to hover above a fixed point on the ground (which is why you can point your TV satellite receiver dish towards a fixed direction in the sky). What is the radius of its orbit? Solution: Using Kepler'; third law again,we get R_(gvo)=x^203=((1)/(273))^2/3=0.110LU=6.65R_(E)approx 42,400km
- 11. A 2.0 kg block is sliding down a 15^circ incline. The coefficient of friction between the block and the incline is 062. At the location shown, the speed of the object is 7.0m/s How much further does the block move before stopping? [5 marks] square square
- Which one of these characteristics does not match with SCIENTIFIC METHOD? process that can only be used in math class to solve geometry Step by step process to approach problems hypothesis experiment
