Answer:
F=2627.6N
Explanation:
The work done by this resistive force while traveling a distance <em>d</em> underwater would be:
where the minus sign appears because the force is upwards and the displacement downwards.
This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:
Putting all together:
Answer:
Fc = [ - 4.45 * 10^-8 j ] N
Explanation:
Given:-
- The masses and the position coordinates from ( 0 , 0 ) are:
Sphere A : ma = 80 kg , ( 0 , 0 )
Sphere B : ma = 60 kg , ( 0.25 , 0 )
Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15
- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2
Find:-
what is the gravitational force on C due to A and B?
Solution:-
- The gravitational force between spheres is given by:
F = G*m1*m2 / r^2
Where, r : The distance between two bodies (sphere).
- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-
Determine the angle (α) between vectors rac and rab using cosine rule:
Determine the angle (β) between vectors rbc and rab using cosine rule:
- Now determine the scalar gravitational forces due to sphere A and B on C:
Between sphere A and C:
Fac = G*ma*mc / rac^2
Fac = (6.674×10−11)*80*0.2 / 0.2^2
Fac = 2.67*10^-8 N
vector Fac = Fac* [ - cos (α) i + - sin (α) j ]
vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]
vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N
Between sphere B and C:
Fbc = G*mb*mc / rbc^2
Fbc = (6.674×10−11)*60*0.2 / 0.15^2
Fbc = 3.56*10^-8 N
vector Fbc = Fbc* [ cos (β) i - sin (β) j ]
vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]
vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N
- The Net gravitational force can now be determined from vector additon of Fac and Fbc:
Fc = vector Fac + vector Fbc
Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8
Fc = [ - 4.45 * 10^-8 j ] N
The initial speed of car A is 15.18 m/s.
Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.
p=p'
pa+pb = pa'+pb'
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)
(1,783×va) + 0 = 14,264+12,800
(1,783×va) = 27,064
va = 15.18 m/s
Learn more about The law of momentum conservation here: brainly.com/question/7538238
#SPJ4