Anjela R.

asked • 11/17/21

Please answer this question correct. i would really appreciate it. Thank you

A child in a boat throws a 5.30-kg package out horizontally with a speed of 10.0 m/s. (Figure 1) Figure1 of 1 Part A Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 27.0 kg and that of the boat is 50.0 kg . (Take the package's direction of motion as positive.)

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William W. answered • 11/17/21

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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

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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 figures
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11/17/21

William W.

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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)
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11/18/21

Anjela R.

Thank you so much !!!
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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= ________ Kg
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11/25/21

William W.

tutor
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 N
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11/25/21

William W.

tutor
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 kg
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11/25/21

Anjela R.

Thank you so much i really appreciate it
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11/27/21

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