\"A Homework 10-Mom ent x ( G The mass of the Earth is 5 × G A 4.30-kg steel bal
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"A Homework 10-Mom ent x ( G The mass of the Earth is 5 × G A 4.30-kg steel ball strike X arcus -> Dwww.webassign.net/web/Student/Assignment-Responses/submit?dep-13131539 A 4.30-kg steel ball strikes a wall with a speed of 11.0 m/s at an angle of = 60.0° with the surface. It bounces off with the same speed and angle (see figure below). If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball (Assume right and up are the positive directions.) Need Help?Read ItMaster It Talk to a Tutor -/1 points SerPSE8 9P036 My Notes Ask Your Teacher The mass of the Earth is 5.98 x 1024 kg, and the mass of the Moon is 7.36 x 1022 kg. The distance of separation, measured between their centers, is 3.84 x 108 m. Locate the center of mass of the system as measured from the center of the Earth Show Ask Your Teacher m from the Earth's center Need Help?Read It Talk to a Tutor 8. 1/2 points | Previous Answers SerPSE8 9.P.037.WI. My Notes Ask Your Teacher Four objects are situated along the y axis as follows: a 2.01-kg object is at +3.05 m, a 3.02-kg object is at +2.53 m, a 2.46-kg object is at the origin, and a 4.05-kg object is at -0.490 m. Where is the center of mass of these objects? X= 3:01 PM 3/23/2016 49.11.Explanation / Answer
Only the x-component of the momentum is changed. The y-component remains the same so Fy = 0
P = m*Vsin = Fx*t
P = Pf - Pi = Fx*t
Pi = (4.30 kg) (11 m/s) sin 60
Pi = 40.96 kg.m/s
Pf =- (4.30 kg) (11 m/s) sin 60
Pf = -40.96 kg.m/s
P = Pf - Pi = -40.96 kg.m/s - 40.96 kg.m/s = -81.91 kg.m/s
so Fx = P / t
Fx = -81.91 / 0.200 = 409.63 N
problem 2
To find a center of gravity you must balance the moment of inertia of each element of the systm.
Moment of inertia = mass*distance to the center of gravity
MeDe = MmDm
De = distance from the Earth to the CG
Dm = distance from the Moon to the CG
De + Dm = Total distance = 3.84 x 10^8 m
Dm = 3.84 x 10^8 - De
Substitute in for Dm...
MeDe = Mm(3.84 x 10^8 - De)
(5.98 x 10^24)De = (7.36 x 10^22)(3.84 x 10^8 - De)
Solve for De...
De = (7.36 x 10^22)(3.84 x 10^8) / (5.98 x 10^24 + 7.36 x 10^22)
De = 4668.69 km
Since the radius of the Earth is about 6371 km, the CG for the system is below the surface of the Earth
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