20814.37/((4/3 *pi)) = 4969.06480926548028
sqrt3(4969.06480926548028) = 17.06442086445121827
2a + 3b sub a = 2-i b = -1+1
2(2-i) + 3 (-1 +i)
4-2i - 3 + 3i
1 + i
First, find the correct graph.... at x = 0
f(x) = -3 g(x) = -3 This conforms to the SECOND graph.....
NOW, f (x) = g(x) where the graphs intersect....which is at x= .....? (two places)
Find the vecor sum of the forces
y direction: 11.8 sin 53.7 + 22.9 sin (-15.8) = 3.27 N
x direction 11.8 cos 53.7 + 22.9 cos(-15.8) = 29.02 N
Net magnitude force = sqrt ( 3.27^2 + 29.02^2) = 29.2 N
F = ma
29.2 = 15.5 a a = 1.88 m/s^2
Make your upload a smaller number of mb.
3.05 x 9.81 N =3.05*9.81 = 29.9205 N (Normal force)
29.9205 N x .250 = 29.9205*.25 = 7.480125 N Frictional force
Hmmmmmm..... I think I agree with Anthrax on this one. The question did not ask for the NORMAL force on the hand....it asks for the magnitude of the NET force on the hand..... Ben, you even stated that the 130 N results in increased frictional FORCES on the hand..... the RESULTANT force acting on the hand is not just the normal force of 1127 , but ALSO the horizontal force of 130 N and these two forces DO add as Anthrax calculated.....and they act in the direction of the hypotenuse (i.e. angled....not NORMAL)..... My one up vote goes with Anthrax.
My 2 cents worth
https://web2.0calc.com/questions/physics_95
Please do not post questions more than once.
The 'net ' forces acting on her are
f = ma
= 57 (9.81 + 1.25) = 11.06*57 = 630.42 N Downward
For the mass of the bird F = .495*9.81 = 4.85595 N Downward so negative - 4.86 N
-4.86 + 7.28 = -4.85595+7.28 = 2.42405 N is the NET force acting UPward on the bird