William W. answered • 11/17/21
Experienced Tutor and Retired Engineer
The momentum at time 1 is zero since initially everything is at rest. At time 2, since momentum is conserved, the momentum must also add to zero. Since the package has a momentum of (5.3)(10.0) = 53.0 kg•m/s, then the boat must have an equal and opposite momentum. The momentum of the boat/child is (27.0 + 50.0)v = 77.0v and that must equal the package momentum
77.0v = 53.0
v = 53.0/77.0 = 0.688 m/s in the opposite direction so since we are told that the package direction is positive, then v = -0.688 m/s
William W.
Initial momentum is (130)(2.1) - (92)(5.3). Final momentum is (130+92)v. So their mutual speed is the absolute value of [(130)(2.1) - (92)(5.3)]/(130+92) = 0.97 (speed would not have a negative sign)11/18/21
Anjela R.
Thank you so much !!!11/18/21
Anjela R.
can you help me with this question i got part A .... A satellite circles a spherical planet of unknown mass in a circular orbit of radius 2.2×10^7 m . The magnitude of the gravitational force exerted on the satellite by the planet is 110 N . What would be the magnitude of the gravitational force exerted on the satellite by the planet if the radius of the orbit were increased to 3.5×107 m ? Express your answer using two significant figures. F = 43 N Part B If the satellite circles the planet once every 2.3 h in the larger orbit, what is the mass of the planet? Express your answer using two significant figures. M= ________ Kg11/25/21
William W.
Part A: Newton's Universal Law of Gravitation says F = (G)(m1)(m2)/r^2 so multiplying both sides by r^2 we get (F)(r^2) = (G)(m1)(m2). We know that "G" is a constant, we also know that the mass of the planet and the mass of the satellite are constant so "(G)(m1)(m2)" is constant. So, in this case (G)(m1)(m2) = (110)(2.2x10^7)^2 = 5.324x10^16. To calculate the force at the new distance, use 5.324x10^16 in place of "(G)(m1)(m2)" in Newton's equation: F = (G)(m1)(m2)/r^2 = (5.324x10^16)/(3.5×10^7)^2 = 43 N11/25/21
William W.
Part B: A variation of Newton's Universal Law of gravitation is the the period (T) can be calculated from T = 2pi[sqrt(r^3/[G*m]) where r is the radius, G is the Gravitational Constant (6.67x10^-11), and m is the mass of the planet. So if we square both sides we get T^2 = 4pi^2(r^3/[G*m]) or m = 4pi^2(r^3)/(G*T^2) but you would need to convert T into seconds (the SI unit of time). T = 2.3*3600 = 8280 seconds therefore m = 4pi^2(3.5x10^7)^3/[(6.67x10^-11)(8280)^2] = 3.7x10^26 kg11/25/21
Anjela R.
Thank you so much i really appreciate it11/27/21
Anjela R.
can you help me with this question too someone else answered this before but it was wrong he got -0.97m/s and i got -1.0 m/s both answers are wrong. A 130-kg tackler moving at 2.1 m/s meets head-on (and tackles) an 92-kg halfback moving at 5.3 m/s . Part A What will be their mutual speed immediately after the collision? Express your answer using two significant figures11/17/21