Question

Im struggling with these two questions if you could please help me with both that would be great!!

1. Previous Answers My Notes 0/2 points Ask Your Teacher SerPSE10 13.1.OP.001. A 260-kg object and a 560-kg object are separated by 4.90 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 30.0-kg object placed midway between them. 1.00e(-7) X N (b) At what position (other than an infinitely remote one) can the 30.0-kg object be placed so as to experience a net force of zero from the other two objects? 1.457 The force due to the larger mass will be greater than the force due to the smaller mass when the distances are equal. Do you expect the object to be closer to the large mass when the forces are equal? m from the 560 kg mass toward the 260 kg mass Need Help? Watch It Read It +Ask Your Teacher 2. Previous Answers My Notes 0/2 points SERPSE10 13.1.OP.003. Three uniform spheres of masses m, = 1.50 kg, m, = 4.00 kg, and m =6.50 kg are placed at the corners of a right triangle (see figure below). Calculate the resultant gravitational force on the object of mass m,, assuming the spheres are isolated from the rest of the Universe. -1.084e-10 X Does the sphere of mass m1 contribute any force in the x direction? î + 4.45e-11 X Remember that superposition allows us to calculate the forces do to each sphere separately and then add the results together. x 10-11 N (0, 3.00) m т (-4.00, 0) m х тz тg (i Need Help? Read It Watch It

Answer 1

Ans! m, ca borg m=560kg m y Cory s (a) from re 4.9 m m₂ = 30kg midway G= 6.674X10" mangis? from Law of garisations we can calculate force bu these objects. . Pie Gaming and r, . r = r2 = 2.45 m 6.674810 " x 260 x 30 2.45*2.45 8.568 Xto -8N 1. fa3 = Gm₂ m₂ 22 6.674x15"x560x30 2.45X2.45 = 18,6798158N Net force on my is f = faz-fiz = 10.1118108 N

for not force to em to o tem m of 06) , be zero. Let us YK→ put the m at r. distance from m. Po fizê faz at 8 Gmim a m.mz (8-82 ma m (8-8)2 82 +0,2–2rr, min 13 . 268 22 +0,2_ 2rr, 28 288 2 = 1382 + 137,2 – 2600 15 r2 = 1382 2600 150, 2 = 1384.9x4.9 - 26x4, 9xr 15,² = 312.13 - 127.42 8,2 + 8.498 - 20.81 - 0

so we solve have a quadratiè in v it - 8.49 + 720801 +33.24 - 8.42 I 12.463 Tora – 10. u4 and 2021 diptance can't be negatie so Ir-2.021 mi

Soin 2 m = listy mamy m2 = 6.5kg ng ale un = & - Agami foa amm 6.674 X 10"x1.5 X Y - 4.45810N. o fa 4 m₂m 22 6.674x15"x6.5x y Resultent - 10.845 x16'' N Resultent force, f = fizh + faz? 19.8025 + 17-614 x16 f . 11.923 x16'' N